Calculus for beginners - Limits of Sequences | Dániel Csíkos | Skillshare

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Calculus for beginners - Limits of Sequences

teacher avatar Dániel Csíkos, Mechanical engineer

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Taught by industry leaders & working professionals
Topics include illustration, design, photography, and more

Lessons in This Class

26 Lessons (1h 41m)
    • 1. Welcome to the course!

      2:22
    • 2. Definition of sequences

      2:19
    • 3. Definition of convergence

      3:32
    • 4. Limits of common sequences

      5:55
    • 5. Necessary condition of convergence

      3:34
    • 6. Fundamental operations

      5:06
    • 7. Order of magnitude

      3:00
    • 8. Limits of fractions - 1. Example

      4:20
    • 9. Limits of fractions - 2. Example

      2:06
    • 10. Limits of fractions - 3. Example

      3:42
    • 11. Limits of fractions - 4. Example

      2:55
    • 12. Limits of fractions - 5. Example

      5:04
    • 13. Searching for index number - 1. Example

      10:06
    • 14. Searching for index number - 2. Example

      6:49
    • 15. Troublesome sequences with square roots - 1. Example

      4:39
    • 16. Troublesome sequences with square roots - 2. Example

      3:00
    • 17. Troublesome sequences with square roots - 3. Example

      3:41
    • 18. Troublesome sequences with cube roots - Example

      5:15
    • 19. Limits related to Euler's number (e) - 1. Example

      3:58
    • 20. Limits related to Euler's number (e) - 2. Example

      4:24
    • 21. Limits related to Euler's number (e) - 3. Example

      3:43
    • 22. Squeeze theorem

      3:00
    • 23. Squeeze theorem - 1. Example

      3:07
    • 24. Squeeze theorem - 2. Example

      2:37
    • 25. Squeeze theorem - 3. Example

      2:41
    • 26. Thank you!

      0:28
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About This Class

This mathematics course has mainly been created for students currently learning calculus at college/university. In this course, you are going to find everything that you need to know about the limits of sequences!

Objective of the Course

The main objective of the course is to help you be able to solve any kind of problems related to the limits of sequences.

What will I learn?

  • How to calculate the limit of sequences
  • Basic notions related to sequences
  • Orders of magnitude of different type of functions
  • Limits of fractions
  • How to determine the index number of convergence
  • How to deal with square roots
  • How to deal with cube roots
  • Definition of Euler's number (e)
  • How to use Euler's number (e) to determine limits
  • Squeeze (Sandwich) theorem

What do I need to know to start the course?

A basic pre-calculus knowledge is enough. The course starts from scratch and take you through the topics with lots of examples.

How to make the most of this course?

There are several practice problems that you can solve by yourself. I suggest solving those problems after watching the lectures of each topic!

Meet Your Teacher

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Dániel Csíkos

Mechanical engineer

Teacher

Welcome! :)

I'm a mechanical engineer and online entrepreneur. I've got my MSc level degree in mechanical engineering at Budapest University of Technology and Economics and I'm enthusiastic about sharing my knowledge and my love of engineering sciences.

I've been teaching and helping mechanical engineering students as a private tutor in various subjects for over 3 years. Therefore I not only know how to understand a topic as a student but also how to make it understandable for others. I passed most of my subjects with an excellent mark and I hope I can help you to reach the desired mark or objective for yourself by ease, too. I offer guidance to understanding either if you are a student or if you want to... See full profile

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Transcripts

1. Welcome to the course!: Welcome to my course. Let me introduce myself on my course about the limits off sequences just in the natural. I am done here, Chico. Sh I'm a mechanical engineer who prefers smiling and injuring life instead of designing in the boring office. I've been in an extraordinary mathematics vacuity in secondary school. I cannot say that I was the best, but I have experienced the best training available. This inspired me to have you. I don't want to let you struggle with mathematics. It is a fine topic, but unfortunately it is not always dealt well. I have several years of teaching experience and I can put myself into your position. So I think I'm Duval. Who can help you? I designed this course for students having trouble with calculating the limits off sequences. I tell you the theory in a nutshell, as I rather wanted. Concentrate on solving problems. Therefore, a show you lots off numerical examples, as I believe that this is a better way of learning any topic. But it is never enough to just see the solution. I would like you to get involved. You are going to get practice problems designed for you. These are going to have to really understand the topic by applying everything you learned. So what is going to be in the course? A short in terra radical introduction about the basics? But sequences are what do I mean by the limit and soul? Nothing to mathematical. Then we jump deep into calculating the limits. First, you can learn how to calculate the limits of fractions. Then you are going to be able to calculate index numbers for describing the convergence. After that, I make you able to handle square and cube roots in trouble, some sequences and actually to hand the roots even in other topics. Then you are going to become an expert off the limits based on E. On. Finally, we discussed the practical applications off this creased era. This is all you need if you want to determine limits off sequences, so join me and let's learn everything in practice 2. Definition of sequences: first, I'm going to introduce the fundamental notions to you. So in this lesson, you are going to get to know what sequences are. The first and most important thing is to know what the sequence meals. It is basically a list of numbers. These numbers are just written in a specific order. This is represented by the notations to AM means the ant term off the sequence where and is the subscript. If I mention Avon, then I think of the first term off the sequence. A two is second. A tree is the third and so technically, you don't have to start the subscript with an A 4 to 1. But still, that is very common. Let me highlight a very important detail. The South script can only be an integer number. Therefore, sequences are really just leased off numbers. They are very different from the continuous functions. Functions can be differentiated or integrated, but there is no way to def initiate the sequence. Just remember that the most important difference between a function on the sequence is that the graph off a function can be continuous. But the graph off a sequence only contains some points like you can see it on the graph. The terms off the sequins are driven by a formula. For example, a equals 21 over. In this case, the ants term off. The sequence can be calculated by this formula. This is an expletive formula, but the Anstrom can be calculated with the half off the subscript that can also be recursive formulas, where you can only calculate the anstrom with previous terms. If we get back to the given sequence, you can write its terms easily. They are Bomb 1/2 1/3 1/4. And so, in case of the 1st 1 you should write one instead off than two for the second. And so it goes. This is a pretty simple sequence, but the main thing is to be able to calculate all of the terms according to the formula. 3. Definition of convergence: after knowing what the sequence is, let's get to know what do we mean by its convergence? When can we say that sequences is convergent? Whenever the secrets hasn't limit? We can say that it is convergent, but we haven't got much further. What do I mean by a limit? The notation for a limit calculation looks like this limb A M equals to hell. Here Lim is a shorter for the Latin Burger Lima's which basically means limit under dilemmas. We indicate that n goes to infinity. This means that the limit is investigated with a condition that must increase until its value goes to infinity. In case of sequences, you can only check the limit at infinity as there is no other meaningful limit. If I am is convergent at infinity, then there is a limit which I indicated with L. This l is a constant number. We would love to know when this limit exists and what is its value? There exists a limit if the inequality absolute value a minus l is smaller than absolute is satisfied whenever or subscript and is greater than the number marked by capital. But does this mean al denoted delimit number. Therefore A and minus L is the difference off the answer. Turn off the sequence and the limit off the sequence. The absolute value of this difference is a positive number, which describes the distance from the limit. Absolutely is a very small number which can be chosen to be any positive rial number. If the limit really exists for every possible absolute, there's a capital and number which you can use in the second inequality were together. The whole sentence means that after a certain term off the sequence, every term off the sequence is within a given distance to the limit. So every term is in the neighborhood off the limit number. As I mentioned in case of forgiven absolute, we can calculate the sufficient value off on you should be able to determine I am, but I'm going to teach you this later. Now you should just memorize that the sequences convergent if after a certain term, all if its terms are as close to the limit as you wish. That was the case when the sequence is convergent. But if it is not convergent, if there is no limit, then we can say that the sequence is divergent. Divergence means the opposite of convergent. Whenever the sequence is not convergent, we can say that it's the virgins. However, there are multiple ways how the sequence can be divergent. I just want to mention want special case. If the limit off a sequence equals two plus or minus infinity, then we can say that the sequence converges the plus or minus infinity or in other works. We can say that the sequence is divergent. This is the only case where both convergent and divergent can be used in other cases. If there is no convergence, then the sequence is just the virgins. 4. Limits of common sequences: let me show you the limits of common sequences. You can use these as a starting point when you face more complex problems. Let's start with simple Polina MIAs. The limit off em at infinity is obviously infinity. The sequence contents the positive integer numbers in increasing order. So the sequence converges to plus infinity. The same goes if any positive power off an is taken and to the power off. K also converges to plus infinity as and goes to infinity to be more specific. This is through off on squared and cubed And so we're so technically scrabbled and is the same as it equals 10 to the power off one over to this is also sequence with increasing terms. On its limit is infinity so whenever and is raised to a positive power. The sequence goes to infinity. But what about negative powers? They mean fractions for example one over and equals to enter the power off minus one. The limit can be returned by the half of the previous results. The nominator goes to one and the denominator goes to infinity. So the whole expression goes to one over infinity. This means zero as one divided by a huge number means a very small number. If one over and to the power of K is taken where K is positive, the same happens the denominator goes to, Infinity said. The fraction goes to zero at infinity. Just keep these in mind when you have to deal with Polina meals or fractions. Let's continue with exponential expressions, for example, two to the power off and goes to infinity if and goes to infinity. The sequence includes the pavers of two, so it starts at two for 8 16 and goes on to infinity. On the other hand, one over to the power off goes to zero. At infinity, the sequence starts with number 2 1/4 and so you can see that the terms are decreasing on. They go to zero or you can think about. The previous result is 1/2 to the power of an it 1st 1/2 to the power off, and you get a result where Von is divided by infinity, which really goes to zero. Let's get the concept. Once off this, the first sequence was divergent and the other was convergent, with the limit off zero so if you take a to the power of our but the absolute value off a is greater down Von. Then the sequence is divergent as taking the father's off a means that the absolute value is increasing with every stuff. If you see the same expression but the absolute value of a is smaller dump on you know, the delimited zero. Just think about this. If you take the paper off a number smoother dumb bombs, you always get a smaller number. You could notice that I haven't included the case, but absolute value off a is bomb. If a equals to one than the sequence contest only wants, the sequence obviously converges to farm. If a is minus one, then you get an alternating sequence. The limit does not exist in this case. Finally, the sequence consists off one and minus monk repeatedly, there are infinite number of ones and minus wants in the sequence. But neither of those numbers can be considered as limits. In case off limits. You can say that they're both a certain term. Every term is within the small neighborhood of the limit. Here. That is just not true. Both one on minus one our so called limit points around which you can find an infinite number of terms. However, the limit points are not necessarily limits to the sequence. There is always only one limited sequence. Besides the powers, it is important to see the roots to. If you get the terms of a sequence by taking the an threw off rial number, the sequence becomes convergent on the LTD's bomb. As you might remember, if you take a greater route, the result gets closer to one. But what if there is something else under the route? The entered off m also goes toe. What if and goes to infinity? This is a great result with which you can start if you face more complex problems. Please remember to this one, there is one more interesting common sequence which has boastful in the mail on exponential origin, the limit off one plus one over and to the power off and is Oilers number E at infinity. This can be proved nicely, but I don't want to bore you with that. Actually, this is the definition off Oilers number, so you can just use this if a similar sequence comes in your way. More generally von plus K over and the power off all times and is also convergent. Its limit is eat to the power off K times l. This could be proved by the help of the previous result. Actually, I'm going to show you this in the sample problems. For now, just know that the similar sequences have limits connected to the Oilers number. 5. Necessary condition of convergence: this lesson is quite heretical. But let me shortly tell you the necessary condition off convergence, which you could use if you cannot relate sequence to a common limit or even worse, you can apply the definition of convergence directly. Fortunately, there is a necessary condition of convergence, which is way easier to use than the definition of convergence by itself. In practice, you usually decide about convergence by knowing some common sequences. But with this condition, you can make decision about any sequences. Convergence a sequence is convergent if it is increasing and it is banded from above, or it is convergent if it is decreasing and it is bounded from below. In the first case, there's an upper limit to which the terms are getting closer and closer as their value is increasing. However, they can never cross that upper limit. As the value is banded from above, you can find the smallest number, which can be used as this upper limit. The sequence is going to converge to death. Upper limit. In practice, you don't have to determine the smallest upper limit to say that the sequences convergent if you can't say on a number, which is greater than the terms of the sequence, and you can also prove that the terms of the sequence on increasing order then you can be sure that the sequence is convergent. The same goes for the second case that is a lower limit on the terms off the sequence are approaching the limit. So actually, the sequence converges to the greatest lower limit. Let me clarify the details. If you want to understand the previous definition, even more monotone E can be defined in two ways. The sequence can be called increasing if given term off a sequence is greater, or at least equal to the previous, so a one plus one is greater or equal to a. This is enough if you investigate convergence. However, if someone says that the sequence is strictly increasing, the condition says more. In this case, the quality is not allowed. I just wanted to mention this to get familiar with the terminology. Also, there are decreasing sequences where a given term off the sequence is always smaller or equal to the previous one. And there are Street Lee decreasing sequences where given term is always smaller than the previous. What does it mean? that the sequence is bounded simply, a sequence is bounded from above. If there exists a number which is greater than every term off the sequence. This is not a very official definition, but this covers the main point in the other direction. A sequence is bounded from below. If there exists a number which is smaller than Africa off the sequence Now that everything is clarified, let me show you the necessary condition of convergence. Once again, a sequence is convergent either if it is increasing on, banded from a both or if it is decreasing and bounded from below. 6. Fundamental operations: it is important to know what kind of operations you can do with sequences. For example, what is the limit of two sequences? If they are at it, let's start by knowing the limit off two sequences. The limit off AM is capital A. On the limit of beyond is Capital B. These sequences are going to be used in the fundamental operations. If you add the sequences by adding each term in the sequences to the corresponding term, in the other secrets, you get a result on sequence. The limit off the some of the sequences is the same as the some off the limits off each sequence. The same goes for that difference. You can just take the difference of the limits of the two individuals sequences. If you multiply every term off the secrets by the same really constant, you get a sequence for which the limit is the constant times the original in it. Naturally, if every term is more deployed, the limit must be multiplied with the same costumes. If you multiply the terms off the sequences to get a new sequence, the resultant limit is simply the product off the limits of the two original sick France's . This knowledge is very useful. You can face the pretty complex product, but if you can tell the limit off each part by itself, you can tell the resultant limit by solving simple problems. Similarly, if you divide the terms off a sequence with the terms in the other sequence, you can also divide the limits to get the limit of the resultant sequence off course. You can never divide by zero, so neither the terms of be an nor capital B can be zero. These four fundamental operations are frequently used in practice, so make sure to apply them. Sometimes it is even better to use the consequences off the previous lows. Let me show you some cases now and later. You are going to see everything in practice at the numerical examples. So let's start. If a sequence with the limit of infinity is given on the terms off. This a and sequence on multiplied within known theory, a constant than the product is also divergent. It's limit is going to be infinity. IFC doesn't think about a zero if you deal with the product from which one part goes to zero and the other part is bounded than the product also goes to zero at infinity. When I say that the sequences banded, just think off any sequence that has affiliate limit. The important thing is that be as limit is not class for minus infinity. In this case, the limit off the product becomes the limit of zero times something which must be zero. If am goes to infinity, you can think in the similar manner if the limit off B is not zero, but the positive number than the limit off the product is infinity. Obviously infinity times a non zero number. It was the plus or minus infinity. If you multiply with positive number, it is plus infinity. If you multiply with a negative number, it is minus infinity. Just simply think of the product of the limits in case of these results, after the product will. Let's see the connections in case of fractions. If am goes to infinity and it is derided with anything different from infinity, the result becomes infinite. This is the case when B M is considered bounded. It's limit is some finished number. If you divide infinity by this finish number, you still have infinity What if a M has a feel it, but no serial limit on being has zero limit. If you divide something with the number close to zero, the result tends to be a great number. If you go closer to zero in the denominator, the result tends to infinity. If the limit off AM is positive, then you get plus infinity. If it had been negative, the result would have bean minus infinity. It is important to remark that we don't divide with zero. We just divide with a number which goes to zero, so it is very close to it, but it does never reach zero importantly, the recipe local off such number tends to infinity. In the last case, not am be bounded on the limit off beyond, to be infinity. If you divide affinity number with an infinite one, the result becomes zero. So this resultant limit is zero. These consequences can easily be putting news in practice. If they are not clear yet, do not worry. It is going to be better after you have seen the sample problems 7. Order of magnitude: it is rooted in the order of magnitude of functions compared to each other. If you do know it, you can easily calculate limits, even if different type of functions are present in the sequence. What do I think off when I say order of magnitude? Even in simple cases, there is a difference in order of magnitude between functions of the same kind. For example, and squared is an order of magnitude greater than why is it important? This determines the limits off that quote. The limit off and scratch over AM is infinite, as inscribed is an order of magnitude. Greater than this can easily be seen language writing the Croce and and see that it equals two. However, there are expressions, but you can't just re bride the quot and so easily, but you still have to determine a limit. Let me just show how the functions relate to each other. I skipped the proof, but each inequality can be proven and to the power of and has the greatest order of magnitude. The next one is on factorial than comes the exponential function, which means a to the power of and where a is greater done von So, for example, and factorial is an order of magnitude greater down two to the power off on or three to the power off. But still, to the power of M is greater than any food in a meal expression like and squared or an cute . If we go on, you can see that the integer pavers off are greater than and squared, which is also power off. And but with a smaller exponent. The lowest order of magnitude belongs to the logarithms. Off on this answer the list. All of these expressions have infinitely midst, but you can put them in order. These relationships are important in practice. You can always use that the expression can be divided with other off over order of magnitude and the fruity and steer tensed. A infinity, for example, and to the power off over an factorial has infinite limit. If you switch the nominator on the denominator, the limit becomes zero. For example, lucrative em over and to the power of K equals 20 Even though Boesel the denominator and nominator tends to infinity and to the power of K is an order of magnitude, Greater said, the result becomes zero. Similarly, you compare any off the functions and see the limit. This is how you prove the relationships between them. But now just enjoy that. You can use these results. 8. Limits of fractions - 1. Example: after getting to know the terror ethical background on the limits of the common sequences. Now it is time to solve all kind of problems that start with the limits off fractions. In the first example, we must determine the limits off the given sequence, the and term off the secrets is to and scratch, plus five and blustery over and cubed plus six and squared plus nine and plus one. The limit is investigated at infinity so and goes to infinity if and goes to infinity, so that's and scrapped. Therefore, the nominator goes to infinity. Similarly, all terms in the denominator go to infinity except for the classroom. Therefore, posted, nominator and denominator are divergent. We must determine the limit off the court and off the expressions that are divergent but do not very. Let's just reform later expression to avoid the infinity over infinity format. There is a basic trick that you can always use when you have to calculate limits of fractions. Just divide each term by the greatest one. The greatest term is meant to be the vault, which is the greatest at infinity, for example, in the present case and Cube is smoother than six and scratch if and equals to one or two. But certainly an cubed is greater for huge values. Off and cube is an order of magnitude greater than anything else in the expression. So we should divide the nominator and the denominator by an cute generally. You just have to look for the greatest exponent in case of poor Nahmias. And then you should just divide with that in practice, boast the nominator and denominator can reformed as an cute times, something the n cubed over and cubed is just bomb. You might not even write it down. The rest of the expression is important. The nominator is too over and plus five over and squared, plus three over and cute, and the denominator is one classic silver and plus nine over and squared, plus one over an cubed. The fracture is exactly the same as before, but it is easier to handle the limits of each term separately. Let's start with the nominator. The first term is to over as and goes to infinity To is divided by a number going to infinity, therefore, to over and boost 01 over and goes to zero. This is even a common limit. The multiplication by two does not change its behavior. The same goes for the other terms to five over and squared also goes to zero. As ants crowd grows to infinity and the same goes for tree over and cute as an cube ghosted infinity that four terms in the nominator go to zero in infinity. Similarly, you can deal with the tribes in the denominator. The terms divided by um and squared or cubed are simply going to zero. So everything goes to zero except for the class bomb. What together the limit of the nominator is zero as it is zero plus zero plus zero. The Dominator is going toe want plus zero plus zero plus zero. So it's limit is one. Finally, the fraction goes to zero were fun, which means zero. You could not this have the formalities, Burke. After the final equation, you can just write a result for the limit without right England. Lima's as it is finally, the number that the sequences converge at infinity before getting the result, it is voted to use arrows to mark that ever each done goes. You can't simply ride that to over Antichrist zero. You can just indicate that it goes to zero if angles to infinity. That is why you can see me using this arrows all the time. 9. Limits of fractions - 2. Example: let's continue to determine limits of fractions. In the second example, the ants term of the investigated sequence is five times into the power of eight plus two times and cute plus 7/7 times. Enter the power off eight plus four times and to the power of five plus nine times and squared plus one. The limit shall be determined if and goes to infinity. To solve this problem, we can apply the things we learned in the previous lecture. First, let's find a term which is going to be the greatest for large and values. The greatest term is an to the power off eight, as that has the greatest exponent. That's divide the denominator and denominator. With this term, the nominator becomes five plus two over entered the power of five plus seven over and to the power off eight. The denominator becomes seven plus four over and cubed, plus nine over and to the power of six. Last one over entered the power of eight. We can just write one instead of enter the father of eight over and to the power of eight, as it is just a formality to write it down after this step. Let's just investigate every term separately as n goes to infinity every term which includes one over and we won over, entered the power off. Something goes to zero at infinity. You can mark this for every term. After this, you can check which terms are not going to zero. The limit off the nominator is going to be five and the limit off the denominator is going to be seven. Therefore, the limit off the fraction is 5/7. All other terms have zero limited infinity, so you don't really have to care about them. You can apply this matter for fractions. It is always going to work. 10. Limits of fractions - 3. Example: let's see 1/3 example, which is similar to the previous ones but alot us to do a slightly different solution. This time we must calculate the limit off three times into the power of four plus two times and squared, plus eight over and cubed plus T one. Let's use the usual mattered and search for the greatest term it is entered the power of for so we can divide everything with an to the power of four and religious solution. After the division, denominator reads as three plus two over and squared, plus eight over an to the power of four on the denominator rates as one over on plus two over and cute. We can simply five multiplier as it equals one anyway. And after this, we can use that the constant divided by am always stance to zero at infinity in the nominator, only the three is not going to attempt to zero. Everything else tends to it in the denominator. Everything just goes to zero at infinity. Therefore, the limit seems to be 3/0. It is not nice to write it down like this instant off that just think about the end result you have a non zero limit for the nominator as it is going to be three. This tree is divided by a number which tends to zero as you divide it, something so small you get a great number as a result of the division, so the whole fraction tends to infinity. It is through, as actually you could already learn it in the heretical lessons. Basically, you can use that 3/0 goes to Infinity Steagall. It is not arrogance to ride 3/0 as a result. That is why I just put it into brackets. Let me show you more arrogant solution, which avoids having a division by zero. Let's just consider that the greatest term in the denominator is an cute let's just divide with distance. This is going to lead to a result where the denominator certainly has a fin it No. Zero limit. After the division, the nominator reads as three and plus two over the last eight over and cute and the denominator reads as one plus do over and squad. You can see that in fact, the denominator does not 10 to 0 as the limit off one is one only the two over and it's clad. Goes to zero in the nominator, the two over and on the eight over and cute goes to zero but three and goes to infinity. You can write this down in the nominator. There is a term going to infinity and you can add zeros to it. The nominator just goes to infinity. The denominator goes to one. So if you divide infinity by from, you are just going to have infinity as an answers art. This arms result is the same as with the other mattered. But maybe the solution is straightforward as we cannot really divide with zero limit according to the original population Those that I showed you before. However, both solutions are correct. You can choose whichever one you prefer. 11. Limits of fractions - 4. Example: after living with Polina meals, it is time to investigate exponential expressions. In this forced example, we must investigate a fraction, including several exponential terms. Three to the power off on plus two to the power of um plus 10 to the power of 100 over five to the power of and plus minders want to. The power of M is the fraction. Basically, we have to do the same as before. Let's look for the greatest girl undivided everything with it. 10 to the power of 100 is quite a huge number, but still it is just a constant, so it cannot be the greatest ERM, the exponential terms quickly become even larger. Let's choose among them, you basically have to look for the exponential with the greatest base that is going to be five to the power off. Um, this is the term with which we must divide everything else after the division, we can Friday terms as we want. It is better to write 3/5 to the power of and instead of just using, treated the proper on over five to the power of Why is it so if you multiply a number smoother than one with itself, you get a smaller number. So if the base of the explanations show is smother damper than the term tends to zero as n goes to infinity, let's check everything term. By term, we don't really need the multiplier in front. Just indicate that divided by five to the power of So it is just 3/5 to the power of an intense 20 as n goes to infinity, as I mentioned on exponential, with the base smoother down, one just goes to zero as it gets smaller and smaller with each multiplication similarly to over five to the power of Angus zero to the minus 1/5 to the power of, um is basically the same case. Don't worry about the minus sign. It just means that the science is always changing as and gets greater. But still the term goes to 0 10 to the power of 100 is a constant number five to the father of AM tends to infinity. So if you divided constant with it, you get a zero result. Therefore, this term also goes to zero. What together? The whole nominator goes to zero. You can only find ones No zero number in the denominator which is one. The result becomes 0/1, which he closed zero. As you could see, the solution matter is the same as in case of pulling a meals. Only the greatest term has a finished limit after the division. So in this way you obviously get a result. 12. Limits of fractions - 5. Example: let me show you how you can use everything that you have learned so far. The fracture and contents pulling no meals and explanation shows 22 to the power off, three out plus one plus three to the power of to one plus two, plus 25 to the power of an over to over nine to the power off am plus and to the power of six times two to the power off and is investigated As n goes to infinity, the solution mattered must be the usual. Let's find the greatest now it is not easy to immediately find the greatest 1 25 seems to be the greatest base off all ex financials, but it's exponent is no the same as the others. This variety everything. It is better to have the same exponent for all expert on short terms that's use the experiment, and for every term we can make some rearrangements to reach that goal. You can see the result, but let me guide you through the steps we can start with due to the power off three and plus one. The plus one is just in a way. Let's make it disappear. The plus one in the experiment with a multiplication with two, we can just write this multiplier down So only two to the father off three and remains It doesn't matter if you take two to the power of three were to the proud father of three and then this to the power of to cute means eight. So that is what you can see. That is how we go two times eight to the power off on and we order the other terms in the same manner treated. The power of to and plus two is the next month. The last two means two multiplication is by three. So basically a multiplication by nine treated the power of to L remains, which equals to nine to the power off. Um as three squad is nine and that is taken to the power of 25 is raised to the power of an over to the division by two means taking the square root. Let's do that first and race to the ants power later. Squarely. 25 minutes five. So that is on the ants power. This has been quite a lot of information, but just keep the behavior off the exponential terms in mind With the bitter fruit in, you can just do this reformulation quite fast. Now you can check for the greatest term as it is going to mean the greatest based off all ex financials. Why don't I care? With an to the power of six? It is just waste Moeller in terms off order of magnitude than the exponential terms. I have shown you the comparision off orders of mine. It'd before. Just think of that. The greatest term is nine to the power off. Um is that has the greatest base? Let's just divide. With this done, the nominator becomes two times 8/9 to the power off one plus nine plus five overnight. The power off the denominator becomes one plus entered the power off six times to over nine . To the power of it's better to ride the terms. In this way, it has to recognize the limits off each expression. The multiplier is just one, so we simplify and then let's see the terms. If something smaller than one is raising the ante, Father, it is going to 10 to 0. Therefore 8/9 to the power off and goes 0 5/9 to the power of Anton's zero and the limit off to over nine to the power off and is also zero. You don't really have to care about the constant where the players, as year old times constant is also zero. However we should talk about and to the power of six. It goes to infinity, so basically there's a multiplication, including on infinite on the zero limit. This time you can just think off the orders of magnitude, as I showed you before. The explanation. Short terms are greater order of magnitude than the polynomial terms, so when you multiplied them, the extra my shirt on dominates the behavior off the product. So the limit of the product is zero. You can always use the orders of magnitude if there are different types of functions. What together the limit of the nominator is not under the limit of the denominator? Respond that, for the limit of the fraction, is nine. That's how you can solve the most complicated case off this kind 13. Searching for index number - 1. Example: Now that you know how to determine the limits of fractions, let's take a step forward on determine an index number for the convergence. Let me show you everything. Through an example, a sequence is given. It's an still is two times and to the power of four minus and cubed plus four and over and to the power of four minus and cubed plus eight. First, the limit shall be determined, and then an index number is needed for which, if N is greater than this index number, the ant term off the sequence is waiting absolute distance to the limit off the sequence. First, just forget the index number and concentrate on calculating the limit. Then I show you how to calculate this in lax number. Let's find the greatest term in the fraction it is under the povero for Let's divide every term with it. After the division, the nominator is two minus one over on plus four over on. Cute on the denominator is one minus one over on plus eight over and to the power of four as usual, Everything which is divided with and or its powers 10 to 0 as N goes to infinity, so the nominator tends to two plus zero plus zero. If you check it turned back to the denominator turns the one minus zero plus zero. The limit becomes to over one, which equals to two. So far, everything has gone as usual. Let's start dealing with the second part of the problem. The limit is needed as an input, so you must determine it, even if it is not a separate question. Actually, the question has a little bit, as a condition is given, with which we can determine an index number. The terms of the sequence are written upside on distance to the limit after this specific index number. This means that after a while, the difference between the terms of the sequence and the limit is smaller than Epsilon. It is easy to write this in a formula. A N is the answer. Turn off the secrets Capital Al is the limit off the secrets which he coasted to in your case. And finally, ab silent is just an arbitrary number, which gives us how close the sequence need to be to its live it. The difference is investigated, but you can see that I use the absolute value off the difference. It doesn't matter whether I am a smaller or larger than the limit. We just need that difference to be smaller than the Given Upsilon. For example, if upside er is 0.1 eight and should be within 1.9 on 2.4, basically, you can just ride this inequality up and saw off this the absolute value. A minus are smaller than epsilon. Expression is a general. Now let's see how we can solve this inequality Mathematical First, it would be nice to know if the expression within the absolute value is positive or negative. Looking at them like this doesn't help me. I'm just clueless. It is better to reformulate. The first idea is to use just one fraction that use a common denominator for the terms two is multiplied and also divided by the denominator off the other fraction. In this way, we can get a common denominator. You can do the multiplication on. Then you can just subtract the second nominator from the first bomb. It is how we get one fraction that we can investigate. This fraction is on cubed plus four and minus 16 over entered the power off four minus and cubed plus eight. Let's see if it is positive or negative because I want to leave the absolute value side. Don't forget that M is a positive integer number. According to the definition, the denominator is certainly positive if an is greater or equal to one, and to the power of four is also greater or equal to an cute. The only case of ECU elated is an equals to one. But in that case, the plus eight ensures that the denominator is positive. Denominator is trickier. It is going to be positive if can is great enough as an cubed and four, and it's positive and they are going to get greater and greater, so eventually they are going to be greater than 16. You should check, which is the first positive integer for which the denominator is positive. By trying, you can easily find that even for an egg first tree, the nominator is already positive. So if an is greater or equal to three, the absolute value off a positive number is taken as both the nominator and the denominator are positive. So if an is greater or equal to three, we can just leave the absolute value sign that's continue without the absolute value in this form that in our politic and very arranged and M can be expressed with the half off Epsilon. However, I don't do this rearrangement right away. Why not? Because, theoretically, I could do it. But there is no way that I could really solve a forced order in creation in practice if you are a genius and you can so just go ahead. If you are just like me, you'd better use an estimation. The task has never said that we should give the smallest possible index number. We were asked to give a suit above, um, about which the sequence is written upside on This danced in the limit. Keep this in mind. You can just use estimations. I want to ensure that this original inequality is through. So that's examine a greater expression on the left hand side. If that is smaller than upside on the origin of on its smaller too. To use an overestimate, the nominator can be overestimated or the denominator can be underestimated. I overestimate the nominator by not subtracting 16. This definitely increases denominator. The denominator is underestimated. If eight is not added as the denominator gets smaller, the whole a fraction gets greater. This expression that we got is simpler than the original, but still it is not the best. What is my problem? There's 1/4 order expression on the first order. Expression to the difference is three. This means that even if I divide everything by, um, I would need to handle 1/3 order in creation to find a proper solution. At maximum, I can handle second order creations, so we still need an underestimation. That's further increase the nominator to make the first order term disappear and cute is certainly greater or equal to an as an is a positive integer, so four can be overestimated by four. An acute. This increases the whole expression. Therefore, if five and cube over entered the power of four months and cute is smaller than Absalon, the original expression is going to be smaller to finally, I have got an inequality which I can solve. This is going to need to a first order. In actuality, that's divide by an cute and solve the inequality. Five over and minus one is smaller than epsilon we can rearrange to get a condition for the original inequality is full field if an is greater than five over upside on past bomb, this is one was the solution. But don't forget that there was a condition which we used to leave the absolute value signed. This inequality is only valid if n is greater or equal to tree. Therefore, the suitable index number is greater than the maximum off. Three on five over Upsilon plus Mom, why have I written this? If five over epsilon plus a bomb is greater than three and end is greater in it, the terms of the sequence are written absolute distance to the limit. If five over epsilon plus bond is smaller than three than it is not certainly enough to say that an is greater than five over upsilon possible. You also need that on this greater or equal to treat to have a valued result in practice, just make sure to use the street ist condition. If you use several conditions during the estimations and the reformulation, then just include all of them here. The maximum off or conditions is what you need it off this maximum you can choose an index number as you wish. That is going to be capital and which is a constant number. Forgiven Upsilon. If small on the sub scraped off, the sequence is greater than this capital and than the terms of the secrets are close enough to the limit. The problem is solved if it is not clear yet, just joined me in the next lesson to see an additional example. 14. Searching for index number - 2. Example: in the previous lecture, I showed you the matter with which you can search for an index number. Now let's see an additional example. The ancestor off the sequence is inscribed minus 100 over three and cute, minus and squared times cosign and squared. We should determine the limit off the fraction on. Then we should given index number, above which Africa is waking up silent distance to the limit. Let's start with the limit. What is the greatest term in the fraction? It is an cute so we can divide everything with an cubed after the division, the nominator is one over and minus 100 over and cubed, and the denominator is three miners, one over and Times Co. Side and squad. The limits off one over on 100 over on cute are obviously zero. As n goes to infinity, let's concentrate on the coast. Sign goes on. Scrat can take on values between zero and bomb. The simple co sign or sign could variety between minus one plus one. But by taking the square the when you can only be positive. The main point is to find that coastline and squared is bonded. It can't go to infinity, and we know that one over Anton's 20 Therefore, their product also tends to zero as the product off affinity number, and zero is always zero. That could only be doubts if Cosan could tend to infinity, but it doesn't do that. So back to the fraction the nominator tends to zero, and it's denominator tested tree. The limit of the sequence becomes zero. Now that we know the limit, we can determine the index number. Let's right the inactive idea. The absolute value off the difference off the ants turn off. The sequence on the limit must be smaller than upside. So we take the ants term off the sequence, which is an squared minus 100 over tree and cubed minus and squared Times co sign and squad the substrate. The limit, which is zero down. We take the absolute value we should check whether there is positive or negative number within the absolute value. We should get one fraction to make. Our decision is your. Unfortunately, there's only one fraction. If the limit had not been zero, we would have needed to bring the terms the common denominator. Now let's change the sign of the fractions. The denominator is certainly positive. On Cube is greater than on squared, as is a positive integer as an squared is more deployed, with the number not greater than one. The product is definitely smaller, even though and cute. And here we have three and cute. Let's check the nominator. If an is great, it is going to be positive. Exactly. If an requested then the nominator is zero and above death, and the nominator is positive. So if an is greater or equal to tell than the absolute value off a nun, negative number is taken as the denominator is positive on the nominator is either zero or positive. Basically, we can't forget about the absolute value just alone. Levering mark. If the expression inside the absolute value is negative, you have to multiply it with minus one. When you leave the absolute value, the absolute value off a negative number always equals minus one times that negative number . Now we have a positive number. It's absolute value is itself, so we must find on value for which an squared minus 100 over three and cubed manage stance four times. School San and squad. It's smaller than Epsilon. If I just rearranged this, I would get Anak quality with third Order as there is and cubed and the constant to. Also it is pretty hard to express l. If there is both and cubed and Coulson and in the same inequality, let's get rid off the coastline and get an inequality. With smaller order, we can overestimate the left hand side. It is pretty easy to overestimate the nominator. Just leave out the minus 100. If we not subtract 100 the nominator gets greater. It's always verse, leaving as much terms as you can. It makes your life easier, especially if you just simply leave them and it leads to overestimation. The denominator is a bit trickier, but if you overestimate co san squad with one, you overestimate the whole fraction. Hussein squad is between zero anvil. If you use Von instead off it than an squared gets a greater multiplier that for a greater number, is subtracted from the denominator on this maze. The denominator smaller if the denominator is smaller than the fraction is greater, so it is in fact, good to use this in estimation. If we just leave the term with and squared out, we would get an underestimation, which is not good. Always make sure to think this through after getting this overestimate. It is pretty simple to express on. With upside on. We can simply five it on squared. Only 1/3 on minus one is left. It must be smaller than upside down. This, in actuality, stands if an is greater than fun over Absalon plus one over tree. Don't forget that this is only true if the absolute value off a known negative numbers taken previously, therefore and should be greater or equal to town. So the suitable index numbers are greater than 10 and greater than one over ab silent plus 1/3. So the maximum off down and one over upsilon plus man over three is suitable. Also, any index numbers are acceptable that are greater than this value for given half signer. You can choose any capital on which satisfies this inequality. Just make sure to use the stricter conditions 15. Troublesome sequences with square roots - 1. Example: let me show you a great matter that you can use when you deal with square root. I would like to show this matter through an example. We should determine the limits off square root to an times Square root three AM, plus two minus square +03 on minus five. If you take a look at this sequence or you can see has known live it square road and tends to infinity as an goes to infinity. As every square root and goes to infinity, so does the ones in the brackets. This is where the problem becomes troublesome. There's a difference off two terms that are divergent, so difference off infinity and infinity should be calculated. We don't have any idea about the limit of the difference in the brackets. You can see that the two terms have the same order of magnitude, so that difference might even be infinite. Or maybe even zero. What if the limited zero This is not necessarily the case, but if the limit would be zero for his difference, then we would have trouble once again scrabbled to an tends to infinity so there could be infinitely time zero, so all together you can see lots of doubts. We certainly have to reformulate this expression. What we can do. Basically all you need is an identity which ellos us to get rid off the square root. This identity is a squared minus. B squared equals two a minus B times a plus B. If you would apply the difference with the some that consists off the same terms, you end up with the scratch terms off the origin of us. In this way, you can really eliminate the square odes. Let's go ahead and do this so we need to multiply. But be careful. If you just multiply, you get a different expression. You also have to divide the product with the multiplier. In this rate, we basically multiplied by farm, so the terms of the secrets do not change. Only their appearance gets to be completely different, so we have a multiplication on the division. We should do the multiplication to get rid of the square roots in the nominator. After the multiplication, we got square root Reon Plus two squared, which is three on plus two, and to get miners square +03 and minus five. Scratch, which is minus three on minus five. Everything is done according to the identity. Finally, we get a much simpler nominator that is going to have affinity limit, or so you can see that the denominator goes to infinity. But you can deal with that later. At the next step, we can just calculate the difference in the nominator off the fraction, and we can also multiply with square it to. Therefore, the nominator becomes seven times square it from here. Everything goes. As you have learned previously, when I should. You have to calculate the limits of fraction. Both the nominator and denominator goes to infinity, but we do not have to worry. That's just the right over terms with the greatest, from the certainly means square root and as that is, the only type of on that appears. So let's divide everything by square root. The nominator becomes seven times square to the denominator becomes square road prey, plus two over on plus harrowed three minus five over. You could notice that the riding with square root and means a division by an under described it. That is why, to on five are divided by now. We can see the limits off each term two over and five over and goes to zero and the other terms that do not include on just go to themselves. Therefore, the limit of the nominator is seven times square it to the limit off the denominator is square root three plus zero plus square root treat minor zero. The result becomes seven times square to over two times scrabbling tree. This matter always works like this. But let me show you somewhere examples to practice. And then I also show you how to deal with travel something glances that include cube roots . 16. Troublesome sequences with square roots - 2. Example: Let's continue to get rid off square. Root the limit off square root for unscarred, plus to a minus square root for inscribed minus and shall be determined at infinity. If you take a closer look, you can see that the limit off both square roots are infinity. Therefore, you should Catlett infinity minus infinity. Once again, we should reform late this expression, the usual identity can be used. A squared minus B squared equals two a minus three times a plus B. A minus B is basically right. You can see in or original sequence. Therefore, we can multiply this with the sound off the same terms as the product equals to the difference off the scratch terms. So let's multiply with square root for and scribe plus two well plus square root for inscribed minus on. But don't forget. Also divide with this. In this way, we basically multiplied by phone. Let's take the productive the expression in the brackets on denominator. The new nominator becomes for all scratched plus two and minus for inscribed minus and if you calculate is difference for and squared drops out and finally expression is not going to be in the form of are the limit would be infinity minus infinity. This is the whole point of the process. The nominator becomes three and after taking the difference, and more importantly, we have got the ancestor off the sequence in the four months where we can apply the previous knowledge about the limits of fractures. Let's find the greatest term in the expression. The nominators greatest term is obviously, um, the denominators. Greatest er under the square root is on squared is and squared is under the square root. It means the same order of magnitude. This time it is basically, um so is the greatest term off both the nominator on the dinner connector. We can divide everything with it on denominator become street and the denominator. We are working under the square root, so we must divide everything with chance Crab. That's how four and squared turns into 42 and there's into two over and minus and turns into minus one over. Now we can change the limit of each term, do over and over and are going to be zero. And there is no other. And in the final expression, therefore, the limit of denominators three on the denominator goes to square root for plus zero plus square root for minus zero, which actually means to Times Square road for as claret. For exhausted to the results, speak first to 3/4. 17. Troublesome sequences with square roots - 3. Example: let me show you even more complicated problem. The answer off the investigated sequence is square root to and plus one minus square, two on minus one over square root and minus square. Root on ministry. We should determine the limit as an goes to infinity. You can see that all of the routes and going to infinity. In this case, we might get appropriate limit after dividing with the greatest term, which is square old. However, after the division, the limits off both the nominator on denominator are zero. Therefore, we will need to decide what the limit 0/0 means. The brother just reformulate the whole expression. Let's get rid of the square. Root off this original expression to be able to take the differences. As usual, we use the identity. A scribe minus B squared equal to a minus B times a plus B as opposed. The nominator and denominator looks like a minus B. We can do the same trick with them. We multiply and divide at the same time with square to one plus one plus square root to one minus one, and also we multiply and divide at the same time with squirreled AM plus Square root on ministry. Let's go on to taking the products. The first fractions nominator should be multiplied by the second fractions. Nominator and the first fractions denominator should be multiplied with the third fractions denominator. In this way, there is no square root in the first fraction. The nominator becomes too, and plus one minus two and minus one, and the denominator becomes on minus and minus tree. The second fractions, nominator and the third fractions denominator become one as we use them at the product. I just put the second and third faction together as we continue the analysis in the first fraction, you can just calculate differences as a consequence and drops out from the fraction only to over three remains from the first fracture. It has really being simplified. Now we should calculate the second fractions limit. It is better than the origin novel, as it is not going to be 0/0. At infinity, we can just divide everything with the greatest term, which is square of. After dividing with scrambled on, denominator becomes one plus square root, one ministry over and and the denominator becomes square root two plus one over and plus square it to mind this fall over three over and fun over. And I'm going to zero as angles to infinity. If we write this down, we basically get the result. The limit Off the second fractions, nominator becomes one plus square root one which equals to two, and the limit off its denominator becomes to times square to so all together, the limit becomes Scrafford to over three. As you can see the matter, Edvard again, even though the original problem was not the same as before, I just ask you to remember that if you face trouble some problems containing square roots, you can always use this mattered. Furthermore, you can do something similar in case of cube roots. Let me assure you that in the next lecture 18. Troublesome sequences with cube roots - Example: let me show you how to have the cube root. If you cannot simply determine the limit of a sequence, this is a harder task than the previous month, but you can always think in the same day. The answer off a sequence is cube root and cubed minus two and minus cube loot and cute plus six. We should determine the limit as N goes to infinity. Both fruits are divergent, so the limit is currently looking as infinity minus infinity. Let's reformulate. However, the cube root is not the same as the square root, which you could easily handle. Steel. Let's use the same tote instead off a minus. B. It would be nice to have them cute to get rid off the cube root. Fortunately, there is a proper turd order identity to do this, a cubed minus be cute equal to a minus. B times a A squared plus a times B plus BCE crab. This is great for us. The original expression can take the roll off a minus bi. We just have to take its product with a proper multiplier, and then we get what we want. We can extend the expression by multiplying and dividing at the same time. This looks much uglier than the process with square roots. But this works too. The roll off A belongs to Cuba loot and cubed minus two well, and the rule of B goes to cube root and cute plus six. Accordingly, to this assignment off rules, we've gotten expression that you can see now we know the identity. So it is a straightforward thing to take the productive, the expression in the brackets on denominator. The result should be the difference off a cute and be cute. According to the rose we assigned and cubed, minus two is a cute and and cute plus six is being cute. The denominator remains quite complicated, but who cares? You should rather concentrate on the nominator. The difference can be calculated and cute drops out. So only minus two and minus six remains. I still don't really care about you formulating the denominator. Let's just try to find the greatest term in the expression that is our main goal. Now. The greatest ERM with nominator is obviously on what is the greatest in the denominator. Cube root and cute equals two. You can see that the Q Boots R squared in the first and third term of the denominator. This means that they are in the same order of magnitude as an squad. The second is going to be also at the same order of mine. It'd. There is an cubed reading. The cube roots subdue Scoob Eudes have the same order of magnitude is on. That product relates to and squared, so the greatest term of the denominator becomes an squad. It is greater than on which was in the nominator. Therefore, we should divide everything with an scrapped after the division, the nominator becomes minus two over on minus six over and scrap. The nominator is a bit more complicated, but I hope you can follow me. The simplest way of thinking is realizing that the division off and squared has been done to eliminate the an cute, which is inside the cube root, which also has been raised. The second power basically just realized that you should divide with n cubed. Inside the cube root, for example, the first term becomes one minus two over and scratch as an cute over an cube is one and two and over, and cute is too over and scratch. Everything is done similarly. Let me show you another way of thinking. We must divide by n squared in the denominator. Basically every cube root is raised the second power. If you want to bring the division inside the scribes term, you only need to derive by. Um Now we just want to divide every cube root by. If you want to bring the division by m inside the cube Ute, you should divide every term by an cubed because the cube root of on cubed is going to be equal to. So that is how you can find out what to derive it. This can be a complicated thing if you have troubles with exponential. But please, Now let me concentrate on determining the limit off the sequence. The main aim was to reach this stage where we can check the limit off each term. Both terms in the nominator 10 to 0 as and goes to infinity. The same goes for everything which is divided by and squatter and cubed. By cracking down or the zeros, you can see what's left. Zero is divided with non zero number, you get zero. As a result, that is the limit of the sequence. In the present case, 19. Limits related to Euler's number (e) - 1. Example: there is a certain group of sequences where the limit can only be determined with the help of the definition off Oilers number. Let me show them to you. Now let's see an example where we should determine the limit off one plus bomb over onto the paper off three and plus two. As I mentioned, the definition off winners number, which can happen this definition is actually a Valium limit. It is known that the limit off one plus one over on to the power of equals to be. If you haven't known this, please memorize this result. If you need to calculate the limits off sequences, this comes up frequently every time. If you see a fraction, including and in its nominator and denominator race to a proper also including on you should use this result together limit. Let's try to find this one plus one over and to the power off in our own sequins. Actually, the format is the important thing. There should be one plus one over something you don't need to care about. This something is the only important thing is to have the same exponent on denominator this time and is the denominator so I want to see and as an exponent to to do that, we should be arranged. The current exponent, which is three and plus to the plus do only means that you should multiply with one plus one over and twice this mood applications can just be done separately and then we can deal with the exponent three. The plus two results in the multiplier one plus one over and to the power of to instead of the exponent three and you can use and as the exponent and you multiplied this one plus one over on to the power of and by itself three times. That is how you get three and as an exponent. And that is why the whole expression is cubed. Actually, all you need is the exponential identities. If you know them, you can easily reformulate this kind of functions. Now we can just go on to check the limits off beach to the limit off 1/0, as usual as and goes to infinity, its reciprocal ghosted zero. So actually the expression inside the brackets goes to Bonn, and also it's scratch version girls to evolve. So the multipliers LTD's one that is always the case if the experiment is a finished number . But if an is the exponent which goes to infinity, the behavior is completely different one plus one over and to the power of and goes to e. But how come that once I say that one plus one over and goes to one so by taking its grabbed expression, steer goes to one? And then I say that by raising it to the ants bother. The resulting expression doesn't go to 11 plus one over and really goes to one. But it is very different to have a fin it or infinite exponent as one plus one over and is slightly greater than it is multiplied by itself. Infinite times. The results can be greater than you don't have to be able to show this, but fortunately we know that one plus one over on to the power of and goes to be, Let's get back to your problem. We actually have the result. It is e to the power of 31 plus one over and to the power of and goes to be it is cute, so the whole expression goes to be cute and then this limit should be multiplied by phone. This is how you can deal with this kind of problems. In the next lectures assure you more challenging examples, so stay with me to practice. 20. Limits related to Euler's number (e) - 2. Example: after the introductory example, let's say something more challenging the limit off an plus several over and minus want to the power of to and sharp determined as angers to infinity. We are currently discussing the topic where every limit is connected to Oilers number. So you have an idea what to do now, However, if you just see an extraction in which a fraction containing on is raised to the power also containing and you should also think off the value limit one plus one over and to the power of and goes to be as n goes to infinity. This is the result which we can use in the solution we should seek for this for months. There is one plus one over something raised to the power identical to the denominator. Let's get this format to get it. We should actually do the division and plus seven should be divided by n minus one as a Pelino Mia. How can we do this? Let's reformulate the nominator to see how it is connected to the denominator, so that's actually rather denominator into the nominator. The main thing is to take care off and in the nominator, the constants doesn't matter at this point. So on plus seven EQ first and minus one plus eight. Now, if you divide and minus one by end minus month, you get one. The other part of the nominator can just remain eight. So we have got one plus a fraction, which is eight over and minus one. Now we are closer to the desired. For there is one plus a fraction. We would like to have fun as denominator of this fracture. To get that, they can just divide the nominator and the denominator by eight. Therefore, the ancestor off the Sikh Vance becomes one plus 1/1 over eight times and minus one to the power of to now, the only one thing is missing. To match the known result, the exponent and the denominator of the fracture must be the same. You can just write the denominator to the exponent and then you have to transform the expression to be equal to the original. You can see the solution if it but let me get it through the steps. As I mentioned, we just write the denominator to the exponent. Then we should express the original to experiment with 1/8 times on minus mom to an equals 2 16 times 1/8 times and minus one plus two 16 times 1/8 times and gives you two down 16 times, 1/8 times minus well gives U minus two, which wasn't there originally. Therefore, to has to be added, so you can see that the experiment 1/8 times and minus one has to be taken 16 times. That is why it is on the 16th power. Still, there remains to multiplier off the same kind. You can just write them separately. That gives you the second term off the product that you can see. Now that we've got the final four months, we can determine the limit. The limit off 1/1 over eight times and minus one is zero. So one plus the zero is going to be vom. If you take the Scrabble, the expression in the brackets, the limit is still Well, what about the other term off the product, you can notice that one plus one over and to the power of am format appears here off course because we wanted to see that this part goes to be according to the van. One result. You can see that this expression is raised the 16th barber. So the result is going to be e to the power of 16. As an expression going toe, he is raised to 16 power and then it is more deployed with something that goes toe one at infinity. This has been quite a task. But you should always find the one plus one over N to the power of and in the original expression and then you are good to go. 21. Limits related to Euler's number (e) - 3. Example: let me show you one more example. Off the limits related to Oilers number. The sequence to investigate is inscribed. Manage one over and squared, plus five to the power off and squared plus three. It is similar to the sequence seen in the previous lecture on the Mattered also remains the same. We know that the limit off one plus one over and to the power of and equals two e Let's get or sequence into this former As a first step, we can express the denominator by the nominator. The nominator can be written as an squared plus 5 +906 This equals two and squared minus mom. But now it is easier to see the result of the division one minus six Over and Squared Plus five is raised to the power off and squared Blustery. We have several options for the next step. That machine approach different from the previous one. Let's get the exponent to be the same as the denominator off the fraction. So the new exponent is n squared plus five as n squared plus five is greater than the original expand. We should divide the new exponential terms with something to get the original expression back as an scratch. Plus three plus two equals two l squared plus five. We need to divide twice with the expression in the brackets. You can check that the resultant expand is really an squared plus three. Let's continue the reformulation. I want to have one in the nominator of the fracture. Furthermore, I wanted to be plus what, as one plus one over and to the power of and is the Sikh fast with no limit. That's just divide denominator and the denominator by minus six. In this way, the new nominator is going to be plus fall in the denominator. Multiplier minus 1/6 appears after we divide by 96. Place the four months off the Expressionist correct with exponent must be reformulated Vance again, the new denominator becomes the new exponent. To have the original expression. This new excellent should be multiplied with minus six. Therefore, the whole exponential terms is raised to the power of minus six. That is how we get the final formative. That sequence, the expression in the scrap brackets are identical to the Valium result. That part is going to be as an approaches infinity this is obvious. This is value made or the effort. This part is raised to the power off minus six, which is going to appear in the result. But first, let's take care of the multiplier. The limit off six over and squared plus five is definitely one As an goes to infinity. If you subtract this from one, the limit is still one. If you raise that to the second proper, the limit is began. Steel Mom, what's so if you take a reciprocal, the limit is far. This is always the case. All of the multipliers that appear because off the compensation of the exponent are going to want as an is going to infinity. The result becomes e to the power off minus six as the first part is raised to the power of minus six. On this E to the power of minus six is multiplied platform. If you have understood this example, you can do basically any kind of problems in this topic 22. Squeeze theorem: this crease theorem or so good sandwich to your arm is a T around, which is very used for them. Practice according to the sandwich to euro, if we know that the limit off A and B and at the same real number. And there's a Turk sequence which is in between them that CNN is also convergent and its limit is the same riel number. I said that CIA is in between the other two sequences. It means that if you compare the terms off, see to the terms off a M, it's every term is greater than the respective term off A. Also, if you made the competition to be on every term of being is greater than the respective terms off. See, therefore an underestimation. Andan overestimation is enough to determine a limit of a sequence. You just need to sequences that have known limits in practice. You should examine the secrets, see if you can't get its limit right away. Then you can use estimations. You can create an underestimation by decreasing the terms of CIA. You just should be careful to choose an underestimation for which you can calculate the limit. Who are so you can choose an overestimation. If the two estimates have the same limit, then you have got the limit off CNN to If the two estimates do not have the same limit. But they have a limit that at least you know that CIA is convergent and its limit is between the limits off the two other sequences. There can also be a little twist on the screens t around. It is possible to use only on underestimation. If I am is smaller than see term barrister and A and is divergent that c and is also divergent. Basically, infinity can be considered as an overestimate for CME. So you get the sandwich to Europe back. This method is good if you suspect that you are dealing with the divergence sequence. If that is your intuition, you can just write an underestimation. There is an other case to it uses only overestimation. If the absolute value off see is smaller than the absolute value of am term by term on the live it off am zero, then a limiter. CIA is also zero. Therefore, if you suspect that the sequence goes to zero at infinity, you can just overestimated with something that also goes to zero. And you have proven your suspect Shin. If you think of a sequence of zeros, you immediately haven't underestimation. So this really means disc Reese the urine to in any form of it. The sandwich dirham is very useful to, so let's go ahead and use it. 23. Squeeze theorem - 1. Example: in this lecture a show you how to apply the scree steer. Um, in practice, the limit off aunt route and plus three is investigated. As on approaches infinity. You cannot get this limit directly. But there are two similar sequences that we can be a doll and through C is going to want mercy marks area constant. This is straightforward. The greater route is taken, the closer the result gets the one It is more complicated to see. But the limit off an through and is also one as an goes to infinity, even though, and would go to infinity. The route dominates the behavior off an Stroot on. So the limit is really bomb. You don't have to prove these. Just use them if you need them. Now we can apply this creased era. We have to under an overestimate the original expression. We have to be careful with the choice of estimations. We should get an easier problem than the original oven, and also the two estimates have to have the same limit. Let's start with the underestimation and threw em is smaller than an Trude on Blustery Mystery is not added to This is a great estimation as the limit off an through and is known it goes to one as angles to infinity. So we have already dealt with the underestimation. Let's see an overestimation and hope that it has the same limit. I have shown you the limits that we know. It would be nice to have only one term under the route to get that we can just overestimate tree by three AM, as is a positive in the jerk tree is definitely greater than three. If is greater than and they are April. If panic first of all this inequality stands for positive integer and values, what together there is four on under the route. It is a simpler expression than the origin offer, but still it's limited. And, um however, it can be further extended to two dudes and through four times and threw em is a better form at the limit of these are already known and through the constant and Andrew and are also going to one as n goes to infinity, as both of them are going to bomb, the product is also going to one, so we have got the limit off the overestimation. Unfortunately, it is the same as the limit of the underestimation. This is where the scree steer misused as the limits of the estimations are the same. The limit of the original sequence is also the same as that should be Between the two estimations, therefore and through on plus, tree also goes too far. This is what we were looking for, Sylvia already. Basically every port in a meal under and fruit has the limit of one. But this is how you come through it. 24. Squeeze theorem - 2. Example: let me show you another example in which the sandwich to your, um, should be applied the limit off and threw four and to the power of five plus seven. Shabby determined. It is still unknown that the aunt route See goes to one. If sees area constant and the limit off Andrew and is also over. Let's look for estimations with which we can use these to find the limit off the original sequence. Let's start with the underestimation. We can leave out the plus seven to decrease the expression, or we could just leave out for and to the power of five. Both of them could work as both of them are rented under estimations off course if we just leave the plus seven out. The estimation is closer to the original expression. However answered. Seven can be 100 more easily. So now I try to use this as an estimation, as seven is a constant number and the untruth of seven miscalculated. The limit is now it is one. So we have got the limit off a sequence smaller than the original. Let's continue with the over estimation and hope that its limit is poor. So von simply, we can overestimate seven by seven times and to the power of five as an is a positive integer. This estimation is valid. What together there is 11 and to the power of five. We don't know the limit directly, but we can use the limits off the terms of a product and through 11 is multiplied by an through and to the power of five, which can be river eaten as an through and to the power of five. Here, I use that it does not matter if you calculate a route on down the power for or in the opposite order. Technically, the exponent is five over. But let's just keep it simple and use this form. It is important that we first take the answered off em because the limit of this expression is now it is one Justus. The limit off an through 11 is far. The limit of the product also becomes one, so the limit off the overestimate is the same as the limit off the underestimate, according to this creased ear, um, as the limits of the estimations are the same, the limit off the original sequence is also the same. Therefore, the limit of the original sequence is over. So far, 25. Squeeze theorem - 3. Example: let me show you more complicated problem. The ants term off the investigated sequences and through tree and squared, minus 1/5 and squared. Plus, we should calculate its limit. As we do not really know it directly, we can turn to the sandwich to your own for help. It is still known that the limits off an through Si and Andrew on our farm these should be used. Let's find appropriate way to underestimate the sequence. There is a fraction under the route. There are various terms until the second order with what if every terms would be off the same kind, for example, only and squared would be in the fraction. Then we could easily simplify the fracture. I made an estimation. According to the previous start, the nominator is underestimated as I leave apart, which would definitely be greater or equal to zero. There was three m squared, look minus one, and I have left out and squad minus mom for any first of one. It zero, but for all other values off and it is positive. So I really left out an expression which is either zero or positive. They left it out from the nominator, which makes the nominator smaller and that makes the fraction smaller to the denominator, must be increased to have a smaller fraction. That is why I overestimated on by inscribed. This is not the only way to underestimate, but it is definitely an underestimation that come work after you have got any kind of underestimation, you can simply fight now. I can just simply fight by an squad in this way only on through to over six remains. The limit of that is one as it is and threw off a constant number. The underestimation is fine. I have got the result. Now let's get an overestimation. My goal is the same as in case off the other estimation. Let's just have second order terms. I increase the nominator by leaving out the minus one. This increases the more fractures. The fracture also gets greater by decreasing the denominator. So I leave out the plus on which really makes the denominators more. I can simplify this estimation. So it is untruth 3/5. Its limit is +12 as the limits of the estimations are the same. The limit of the original sequence is also the same, so it equals to want to 26. Thank you!: Congratulations. You have reached the end of the scores. Thank you for your attention and for staying with me. Ta the end of the course. I'm happy that you choose to learn with me. And I hope you've enjoyed the lectures. If you haven't completed the class project yet, go ahead and try to solve the problems on your own. If you find a challenging task, you can share it with the others. We can solve it together. See you around.