 Hi, I'm Zor. Welcome to Unizor Education. Today we're continuing talking about limits, sequence limits. That's a preparation for limits for functions, and that is a preparation for derivatives. So basically it's all gearing towards derivatives. This lecture and the whole course actually is presented on Unizor.com. I do recommend you to take the whole course and for these lectures go straight to the website because any lecture contains on this website very detailed notes and they're very useful actually to examine before or after the lecture. In addition registered students can take exams on the website. The website is free so it's very useful tool for education. Alright, back to limits. Now before we didn't really concern ourselves too much about difficulties of identifying what is the limit. Our simple examples were really trivial like what's the limit of, I don't know, a sequence something like 1 over n. I mean obviously it's infinitesimal as n goes to infinity goes to zero. Today I would like to talk about cases which are not obvious and in limit theories they usually called indeterminate. So there are many problems where limits are not so obvious as this one and they require certain preparation job from you to determine what exactly is the limit. And what's very important and probably the most important when we will talk about derivatives, all the derivatives are based on taking the limits where it's not obvious basically. For instance in this case we have a constant divided by infinitely growing variable so that's obvious. But if you have one let's say infinitesimal divided by another infinitesimal the result is not so obvious or you subtract one infinitely growing variable from another. Again the answer is supposed to be somehow derived based on certain transformations which you have to do with the sequences in question before it becomes obvious what's the limit actually, what the limit actually is. So today we will talk about these cases and let me just repeat that for derivatives it's exactly these indeterminate cases which occur. All right so here is what we are going to do right now. I'll just give you a few examples of the types of indeterminate forms these sequences might actually take and in each case I will just explain as an example what can be done about this particular case to resolve the limit. And there is no like general approach how to take limits like for any cases but we will discover certain commonalities between whatever the examples I have. All right so let's do it one by one. The first not so simple limit I would like to take is the following. I have a sequence which is a product of infinitesimal and some kind of infinitely growing. So one thing is decreasing as n goes to infinity and another thing is increasing as n goes to infinity. Question is how can I take this limit how do I know what actually is the result of the limit of this particular sequence. Well this is a very very simple case of indeterminate limit because what actually this is it's 2n plus 3 divided by n which is sequence of let's divide it 2 plus 3 over n right. So now we have a constant plus infinitesimal and we know that limit of sum is equal to sum of limits. Now this is the constant so it limit its limit is 2. This has a limit 0 because n is in denominator so the result will be 2. So this particular sequence converges to 2 and as you see we needed to do some very very simple transformation to obtain this. It's not obvious just looking from looking at the original segments what exactly the result is. Now as from the type of indeterminate limit which we are talking about you see this is infinitesimal and this is infinitely growing. So there is some kind of a symbolic determination if you wish or certain label which we can put on the whole class of limits of that type. It's 0 times infinity. 0 signifies infinitesimal and infinity signifies the infinitely growing. So we are resolving in this particular case this type of indeterminism. Now can all limits of this type can be resolved so easily absolutely not and sometimes we might actually have the situation when there is no limit but in this case it's simple. So for each particular case I will present an example of 2 and just to give you a flavor that there is some work which needs to be done to determine what exactly is the limit. Another example is basically based on this one which we have already just considered. Example of which can be symbolically designated this type of symbol infinity divided by infinity. What actually it means I have infinitely growing divided by another let's say infinitely growing sequence. We cannot apply some theorems that the limit of a ratio is equal to ratio of the limits because in this case both numerator and denominator do not have a concrete limit. Both are infinitely growing but at the same time we can make certain modifications transformations of this without basically equivalent transformation without changing anything. So as you understand I can represent numerator as 2n plus 2 plus 1 and divide 2n plus 2 by n plus 1 and I will have 2 plus 1 over n plus 1 right 2n plus 2 and plus 1 that's plus 3. Now and this obviously is converging to 2 because it's 2 plus infinitesimal. So that's just another infinity over infinity. Again it's just a symbol it doesn't really mean the division or anything like that. Just a symbolical designation. So the first one was zero times infinity this one divided by infinity. Okay I think these symbols like infinity over infinity again it's only symbols do not take them as a real operation. Okay what's next? Next is the following. I have sine n divided by n square. Now same thing we cannot apply limit of the ratio is equal to ratio of limit because the top doesn't have a limit at all because it's just a sine it goes from plus 1 to minus 1 so it doesn't have any limits. The bottom is infinitely growing. However we can resolve this limit. How can we do it? Well if you remember and if you don't you can always refer to there was one of the previous lectures where I have actually proven that if you have certain sequence and you know that this sequence that each element of this sequence is between other sequences and you know that both boundaries are going to the same limit then this one also goes to the same limit. So we are basically squeezing this sequence from left and from right and as n goes to infinity if these two converged one in the same number this must actually also converge this number. Now so we can also use it here it's 1 over n square and minus 1 over n square right because the sine is always between minus 1 and 1. Now this one is infinitesimal and this one is infinitesimal because it's a constant divided by n square and then square goes to infinity right. So both of them are converging to zero and that's why this one also converges to zero. So we have the situation here when we have a no non-convergent in the numerator and infinitely growing in the denominator. So that's another case which I wanted to. Okay few more examples. Okay I have here example zero over zero it means one infinitesimal in the numerator and another infinitesimal in the denominator and I would like actually to tell you that this is probably the most frequently occurring type of indeterminate limits which occurs in derivatives because all derivatives are built around this ratio of one infinitesimal over another infinitesimal. So it's a very important case. So let me just make a concrete example. I have 2 to the power of minus n plus 1 plus 3 to the power of minus n divided by 2 to the power of minus n. Okay on the top I have one infinitesimal right. This is negative which means it's 1 over 2 to the power of n to the power of n plus 1. Obviously as n growing 1 over 2 goes to zero. Same thing here. This is 1 over 3 to the power of n. 3 to the power of n is infinitely growing. 1 over 3 to the power of n is infinitesimal. So this is infinitesimal plus this is infinitesimal. Some is infinitesimal and in the bottom also have infinitesimal. So how can I resolve this? Well in this case it's easy. You just divide this by this. So what do we have? 2. When you divide you know that you have to subtract the exponential part right. So it's minus n minus 1 that's this one and you have to subtract minus n which means plus n. That's division this over this. Now division plus this over this it's 3 over 2 to the power of minus n right. Now what is this minus n plus n? So it's 2 to the power of minus 1 which is 1 half plus 3 seconds to the power of minus n. Now this obviously is infinitesimal so this thing converges to 1 half. Okay next. Sine square of 1 over n divided by 1 minus cosine of 1 over n. Okay how can I determine the limit of this one? Well in this case what I would do well obviously as n goes to infinity 1 over n goes to 0 cosine of 1 over n goes to cosine of 0 which is 1. 1 minus 1 is 0 so here we have infinitesimal. Same thing here. As n goes to infinity 1 over n goes to 0, sine of 0 is 0. So we have one infinitesimal divided another 0 over 0 type of indeterminant. Now how can I get rid of this indeterminism? How can I determine? Well in this particular case and in many other obviously you notice that if you will multiply it by 1 over by 1 plus cosine of 1 n here and here what happens? Well this is 1 minus cosine square which is 1 minus cosine square is a sine square right? And sine square of 1 n will will cancel this one and we will have 1 plus cosine 1 n. Alright? So 1 minus cosine square of 1 n which is sine square it will cancel this one and that's the only thing which will be remaining. And the result is what does it converge to? n goes to infinity 1 n goes to 0 so cosine of 1 n goes to cosine of 0 which is 1 so 1 plus 1 is 2. So the limit of this is 2. Now this is a little trick if you wish and I don't think I can you know teach you all the tricks but the more problems like this you will solve obviously the more tricks you will have in your repertoire. But this is just kind of a thing which is obvious if you look at this 1 minus cosine and then it's kind of obvious you have to multiply by 1 plus cosine to get 1 minus cosine square and it will be cancelled with sine square. Okay? Well I'm saying it's obvious because I have already solved it right? For people who didn't it's not so obvious so you have to really think about this and as a result the whole thing actually the whole course is supposed to just cause people thinking so it's obvious. Now next next I have another example of infinitesimal times infinitely growing. Okay 1 over n plus 1 that's my infinitesimal times n square that's my infinitely growing so this is a sequence. How can I find the limit of this thing? Well the easiest way I believe is to replace it with 1 n plus 1 n square minus 1 plus 1. Now n minus 1 is as n square minus 1 is n minus 1 times n plus 1 n plus 1 right? So this times this would give you n minus 1 and this plus this would give you plus this. Now this is much easier because this is infinitely growing plus infinitesimal. Well obviously if you will add infinitesimal to infinite growing you will get still infinitely growing so this thing is infinitely growing to infinity that's basically the result. Now it's not obvious looking just from this that this is a result. I mean it's obvious for some people but they already have this mentally made these kind of manipulations in mind. Another more I would say typical way in this case is represented differently. It's n square divided by and I will take n outside of this so n plus 1 I will represent this. Now I can cancel one of them. Now this is infinitesimal 1 plus infinitesimal is obviously converging to 1 sequence so there is nothing wrong with this and plane n is obviously infinitely growing. So you have infinitely growing divided by something which is very close to 1 which means it doesn't really change the infinite growth phenomenon. Alright so that's one of them. Another is more interesting. Sine 1 over n times n. Now n goes to infinity 1 over n goes to 0, sine of 0 is 0 so this is infinitely infinitesimal and this one is infinitely growing so again we have zero times infinity indeterminate. Now here I would like actually to again go back in memory and in one of the lectures where I was using trigonometry and geometry and the limit theory I actually proved that sine of x divided by x as angle x goes to 0 in regions that this thing is converging to 1. So for the proof of this I can refer you to one of the lectures in trigonometry. In notes for these lectures I actually specify exactly the name of that lecture. Right now I don't remember exactly but something like trigonometry chapter and then using trigonometry for geometrical problems. So I actually have proven this. Now I can use this in this particular case because this is not other but sine 1n divided by 1n. Right? Multiplying by n I replaced with division by 1 over n. Now there is nothing wrong by the way I was using n in denominator because n is never equal to 0 and goes to infinity so we are absolutely fine making these manipulations we are not like losing anything. And this is exactly what this is all about because as n goes to infinity 1 over n goes to 0 that's exactly corresponds to this. So the limit of this is 1. It converges to 1. And by the way the proof of this is quite interesting. I do recommend you to go back and examine this particular lecture. In notes for this lecture I have exact reference to this. Okay, next. Next is when I have infinity over infinity. So I have infinitely growing divided by another infinitely growing. So this sequence. Well this is an example of a sequence which is a ratio of two polynomials. Now my next lecture will be completely dedicated to this type of infinity over infinity limits. Polynomial over polynomial. But in this particular case there is a little bit again a small trick. You obviously notice that some of these coefficients is equal to 0 which means n is equal to 1 must be the solution of the equation when n square plus n minus 2 is equal to 0 which means we can divide it by n minus 1 and represent it as n minus 1 times n plus 2. Now same thing here. You see sum of coefficients is 0 which means if n is equal to 0 the result will be 0. So it also supposed to be factored out n minus 1 and here I will have 2n what plus 5 plus 5. And obviously you can cancel n minus 1 again. n is growing to infinity number so there is no problem of this thing is equal to 0. And this one is a little bit easier because it's linear rather than quadratic. And you can always say that this is n plus 2 divided by 2n plus 5 seconds. Now this can be 5 seconds minus 1 second rate. And now we can put parenthesis here which means this divided by this is 1 half and this divided by this will be 1 minus 2 n plus 5 seconds. Now this is infinitely small. So it's infinitesimal and this is a constant 1 half. So the whole thing would be going to 1 half. It converges to 1 half which most of the people with a little experience would guess just looking at the top coefficients at n square. See 1 half. But we will talk about this in the next lecture which is dedicated to ratio of polynomials. Anything else? Okay here is another one. n square sine 1 over n plus 1 over n. Well that's kind of easy because I can just divide it so I will have n square divided by n I will have n sine 1 over n plus 1 over n right? Now this we have we have already established converges to 1. That's my previous problem and this is infinitesimal so it goes to 0 the result will be 1. So this thing is converging to 1. And the last type of indeterminate forms of limits is infinity minus infinity. And as an example I have this square root n square plus n minus n minus n. That's it. Now whenever you have something like this infinity minus infinity how can you get rid of this indeterminism? Well in this particular case I think it's very conveniently to multiply it by this plus this. So again we were just using it in the previous problem a b times a plus b is equal to 8 square minus b square. And why do I have to do it? Because square of this minus square of this you see n square here and n square here it will cancel out. So the whole thing is equal to so I multiply it by their sum and divide by sum. Now this is square of this minus square of this which is plain n. This is square root of n square plus n plus n. Now did it make our life easier? Yes actually much easier. Because what we can do is we can divide by n numerator and denominator. I will have one here. I will have this one. I will have plus one and my square root would be if I divided by n it means that I can divide square root by n square right? So it would be n square plus n divided by n square. And what is this? Well obviously this is n square over n square it's one plus n over n square is one. And this is now easy because only we have here is one infinitesimal and as n goes to infinity obviously this is the constant this is the constant this is the constant and there is only one infinitesimal here so I will have basically this thing goes to square root of one which is one plus one that's two so it's one half. And another last example on the same type of infinity minus infinity the following n plus one minus logarithm by base two two to the power of n minus one. Okay how can we transform this? Now this is infinitely growing and this is infinitely growing that's why it's infinity minus infinity type of the limit. So somehow we have to make it better. Well how can we make it better? Well here is one of the examples. I can have n, n is equal to logarithm to the power of n by the power of two right? That's the definition of the logarithm. So instead of n I will put so one is still here now log two to the power of n minus log two to the power of n minus one. Now difference between logarithms with the same base as you remember is logarithm of the ratio. Now here it's much easier because this is minus one plus one and I divide this by this and I have one plus log two of one plus one over two to the power of n minus one. And now this is easy because as n goes to infinity this is infinitesimal so one plus infinitesimal so that's very much close to one it's converging to one logarithm of one is zero so one plus so the whole thing converges to one. Again a little transformation is needed to get rid of this indeterminism and basically this is the whole story about indeterminate. I cannot give you the recipe how to get rid of things like this or zero times infinity or zero divided by zero or infinity divided by zero. In every case you have to really think about it so it's a little bit of an art if you wish but again as I was saying before the whole course is supposed to bring you closer to thinking about how to solve the problem. It's not interesting just to use the ready to ready to use recipe which somebody else has discovered that's just you know training your memory which is important but it's not the purpose. The purpose of this is for you to discover for yourself all these little tricks little transformations because the more transformations you will do right now based on your own feelings about how you feel your way in this forest of problems the easier for you would be in the real life to solve the problems which any profession will present to you. Well that's it was this very nice conclusion I will end this lecture. Thank you very much and good luck.