 Hi, I'm Zor. Welcome to Indizor Education. I would like to spend some time just doing some very simple exercises with limits. I have divided it in three different lectures. Today will be the limits as argument goes to infinity, next lecture will be minus negative infinity and then some kind of a finite number as a limit point for an argument. I just decided to increase the number of exercises and that's why I decided to divide it in this way. It doesn't really matter. All right, so this lecture is part of the advanced course of mathematics for teenagers and high school students. You can view this lecture from Unizor.com and I suggest you actually to do all your viewing from this website because it contains detailed notes for each lecture plus there is certain educational functionality like you can enroll into a course or any part of the course like any particular topic, take exam for instance. So it's all useful for good education. So, exercises. I have six different problems and let me just go one by one, couple of easy ones, couple of more difficult. Okay, first one is, so we are talking about x going to infinity. I'm talking about obviously positive infinity. If I don't mention positive or negative, positive is assumed. All right, now my first function is 2x square plus 3x plus 4 divided by 3x square plus 4x plus 5. Well, basically my first three problems are about polynomials and I have the ratio of two polynomials and my basic principle which I would like actually you to know is that forget about all the lower members of the polynomials whenever you have something like this polynomial divided by another polynomial and argument goes to infinity. Look only for the highest powers. So basically in this case the highest powers are the same and the limit would be one with two thirds, the ratio between these coefficients and here is why. Now let's divide everything by x square on the top and x square on the bottom. I mean in fact they're out x square. So it will be x square of 2 plus 3 divided by x plus 4 divided by x square over x square once more, 3 plus 4 over x plus 5 over x square. Now x is going to infinity so it's definitely not equal to 0. That's why we can reduce it and here we have basically we don't have indeterminate infinity over infinity was was indeterminate limit but now we don't have this indeterminate limit because every piece of this as x goes to positive infinity goes to 0. So what's left? Left is two thirds. Now that's very much dependent on the highest coefficients to be of the same power. That's why we have this. What if they're not of the same power? Well there are two cases. One case is this when the numerator has a greater power than denominator. Now what can we do in this case? Well let's do exactly the same thing. x square 2x plus 3 over x plus 4 over x square x square 3 plus 4 over x plus 5 over x square. Now there is absolutely no problems with denominator as x goes to infinity as it goes to 3 and numerator as x goes to infinity this is 0 this is 0 and this goes to positive infinity. So we have positive infinity numerator and bounded value around 3 in the denominator. So what will be the result? Well infinitely increasing divided by something which is around 3 the bounded obviously it will be in infinite increasing so that goes to infinity plus infinity. So when you have the power of the numerator greater than the power of denominator you will always have this result regardless of these smaller powers. And last when we have the other way around if I have 2 here and 3 here and 3 here. Well again x square out x square out we have 2 plus 3 over x plus 4 over x square we have 3x plus 4 over x plus 5 over x square. And now what we have? We have a bounded around 2 in the numerator because these two are infinitely small numbers as x goes to infinity and here we have infinitely increasing these are go to 0. So we have somewhere around 2 divided by infinitely increasing obviously the result would be infinitely decreasing to 0 value because denominator goes to infinity. So again it's because denominator has a higher power than numerator. So whenever you have a ratio of two polynomials and you're talking about argument going to positive infinity all you have to do is compare the highest powers. If numerator is the same as denominator then the result would be just the ratio of their coefficients at these highest powers. If numerator is greater than power numerator is greater than the power of denominator it will go to positive infinity and if denominator is greater it would be going to 0. Alright so these are simple three simple cases. Now something which is also simple but you have to just make a little trick. Actually I think I did explain this before so as x goes to infinity what happens? This is the infinitely small value so it goes to 0 and the sign of something which is changing to 0 would tend to 0 remember this. Now this on the other hand goes to infinity infinitely increasing. So it's kind of an indeterminate infinity times 0. However you can recognize this as this so instead of multiplying by x I divided by 1 over x. Now what do I have? Let's consider two functions. The first function is 1 over x. The second function is sin x over x. As x goes to 0 g of x goes to 1. We know that. It's one of the amazing limits which we were touching in the previous lecture. Now as x goes to infinity this is going to 0 as x goes to infinity. Now we can apply the theorem about compound functions because right now what we have here is a g of f of x. So instead of x we substitute 1 over x and we know that this function goes to whatever the limit of this function is. You see this 0 corresponds to this 0. So that's why the limit of this is equal to 1 as x goes to infinity. So as x goes to infinity f of x goes to 0. And now if argument of g goes to 0 we know that g goes to 1. So the limit is equal to 1. So we're just applying the known limit and the theorem of compounded functions. Now the last couple of examples might be a little bit more interesting because that requires certain, well a little trick maybe if you wish. Okay x divided by 3 to the power of x. Now I'm pretty sure that everybody feels that x which is this, 3 over x is something like this. Obviously 3 over x grows significantly faster, at least a little bit further down the line than function x. Function x grows linearly and 3 to the power of x goes exponential. And we all know that exponential is growing faster than the linear function. I mean there are certain cases for instance you have linear function which is very very steep and exponential is very very very slow. Like this is for instance the curve as our money are growing in the bank. If bank is paying minuscule percentage. However, eventually this thing will always overtake however steep the linear function will be and we will be proving this actually. So in any case right now we have a simple case x over 3 to the power of x and we have to prove that 3 to the power of x really grows significantly faster and even grows faster as x goes to infinity then the whole thing should actually go to zero. I mean that's our natural kind of a guess. Right? Question is how we can prove it. Well we do know certain properties of functions including exponential functions and that's what I'm going to to do right now. So what I will do first my plan is first I would like to prove that x is smaller than 2 to the power of x. Now why do I need it? Well because I cannot really divide x by 3 to the power of x and get some analytical calculations because they're not mixing this is linear function this is exponential function these are not mixing. But if I will prove this then x divided by 3 to the power of x would be less than 2 to the x divided by 3 to the x which is two thirds to the power of x and we know that if this is a smaller than one base for exponential function and this is smaller than one then it will go to zero as x goes to infinity. It's just one of the properties of exponential functions. And by the way I might actually remind you that there is a very nice lecture actually couple of lectures on exponential functions in this course in the algebra chapter or part of this course. Alright so that's my plan. Now if I will do this if I will prove this then this immediately follows which is equal to this which is basically obvious that this will go it's just straight from exponentiality from the properties of exponential function with the base less than one. So that's basically the only thing which I need to prove rigorously as much as possible rigorously. I mean there are always kind of limits to rigorousness. So let me concentrate on this because this will follow basically automatically. How can I do that? Well again I mean if you will try to solve this equation for instance it's not easy because again these are two different functions and they're not mixing together. We know how to solve linear equations. We know how to solve quadratic equations but we don't know how to solve these mixtures right? So we have to do something else and here is what I suggest. Look at the function y is equal to x. Well it's a linear function and it's monotonically increasing. Now if x is natural number nn plus 1 we know basically what x is and we know what 2 to the power of x actually is. But what is x is in between? Then 2 to the power of x is kind of difficult thing. Rational exponents it needs certain explanations. But here is what we do know. We know that this function is monotonically increasing which means that we always have such an equation. x is always between two integer numbers. Right? Now 2 to the power of x is also monotonically increasing and that's one of the properties which we have proven actually again in the course of algebra when I was talking about exponential functions. So 2 to the power of x is as monotonic function also between this and this. If x is between n and n minus 1. So what I'm going to prove, I'm going to prove this. That n plus 1 less than or equal to 2 to the power of n. But I will prove it for any n. Well greater than maybe some 1 or 2 or whatever after some beginning. And if I will prove this then from this this would obviously be less than this. Because this is less than this and this is greater than this. Right? So that's obviously would be the case. So that's all I would like to do right now to prove this for integer n's. Now for integer it's easier. Why it's easier? Because I can use the method of mathematical induction. So let me just do it and this is and this is trivial part obviously. How can I prove that n plus 1 is less than 2 to the power of n? Well let's just check it out. For n is equal to 1 this is 2 and this is 2 to the power of 1 which is 2. But for n it's equal. But for n is equal to 2 this is 3 this is 4. 3 less than 4. For n is equal to 3 it's 4, 8 etc. Obviously this grows much faster. But let's prove it. How can we prove it? Well again using the method of mathematical induction. First for n is equal to 2 it's true because it's 3 and this is 4. Let's assume that for certain n equals to k k plus 1 is less than 2 to the power of k and now we will put k plus 1. Now what do we have to prove? We have to prove that that n plus 1 which is k plus 1 plus 1 less than 2 to the power of k plus 1. But I already know that this by my induction assumption is less than 2 to the power of k plus 1. Now and obviously this is less than 2 to the power of k plus 2 to the power of k. Right? Because 2 to the power of k is greater than 1 and this is 2 to the power of k plus 1. Which is what we actually have to prove. For n is equal to k plus 1 this is k plus 2 and this is 2 to the power of k plus 1. So that's easy. Basically that's it. That's a complete proof of this inequality. And from this inequality as I was just saying before immediately follows that this goes to zero because x over 3 to the power of x is smaller than two-thirds in the power of x and that's obviously goes to zero. By the way I didn't mention it but it's kind of I assume that you understand. You see since x to the 3x is greater than zero and less than 2 to the power of x over 3 to the power of x which is equal to two-thirds of the power of x and this goes to zero. This is again a typical example of the squeeze theorem. That's why this also goes to zero or two policemen in a drunk. Alright so that's relatively easy and very simple kind of usage of monotonicity and induction. So using the monotonicity of these functions we have basically reduced the problem to only integer argument and for integer argument we could do it by induction. Let's do something similar right now. Let's prove that ax plus b would be less than 2 to the power of x and let's consider that a is greater than zero just for it. Okay so we have proved that x is less than 2 to the power of x. Let's just make more general thing. Can that be proven? Yes I just mentioned before that even if it's very steep when a is really very large and even in this case we eventually overcome no matter how steep this linear function is. How can I prove that? Well again absolutely similarly we know that ax plus b is a linear function so it's monotonic. So obviously we can always find for any x something like this where it's in between two integer variable. So ax plus b would be less or equal to a n plus 1 plus b or if you wish a n plus b plus a plus a which is actually a n plus c if you wish. If b plus c is equal to c it's still a constant right? So I will prove that for any constant this would be less than this. Not immediately not for n is equal to 1 but starting at certain point which obviously depends on how steep my a actually is because if a is very large then it will go all the way up right? And 2 to the power of n in the beginning it's not really so steep it goes steeper and steeper and eventually it will overcome. So anyway let's just prove it by induction. So first let's try to find where exactly my first beginning I cannot start from n is equal to 1. I should start from something else. So what should I start it from? Well obviously we can start it from something like this. a n plus c would be less than a plus c times n right? n is natural number so here this is a n plus c n and this is plus c but n is natural so obviously the right sin is greater. Now how can I find now let's choose n is equal to what should it be equal to so that 2 to the power of n would be greater than this. Let's just think about it yeah something like this this is less than than a plus c 2 to the power of n right? Because we have already proven before that n is less than 2 to the power of n. So now what I have to do is something like n should be equal to no that's probably not a good idea that's too much. So what should we start it with so the 2 to the power of n should be greater than a a plus c a plus c n well oh here it is. So let's divide it by 2 to the power of n so I will have one greater than a plus c n over 2 to the power of n let's divide by a plus c n to the power of n less than 1 over a plus c okay here is what I can say we have already proven that n over 2 to the power of n goes to zero it's infinitely small so eventually it will be less than a 1 over a plus c and stay that way so for any epsilon if you wish we can find such delta if you remember that it will be less than this so this would be our epsilon so there is always some kind of a number n after which it would be smaller than epsilon so that particular number of n which we know exists because we have proven that this goes to zero this n would be our beginning so our first statement okay let's put n equals to some initial value from this initial value on we will have this particular inequality true so this is our first step n is equal to n zero which depends on now we were talking about what we're talking about a x plus b should be less than 2 to the power of x right so we have found based on a and b we have found some kind of a number and zero from which we can start already thinking about going forward so graphically this is my a x plus zero this is my 2 to the power of n somewhere they will cross and starting from this point my 2 to the power of x would be greater so this is the point where I start we know it exists because we have already proven that n over 2 to the power of n goes to zero now let's apply our induction again so let's assume that n is equal to k a k plus b less than 2 to the power of k now let's set n is equal to k plus 1 so what do we have we have a k plus 1 plus b it should be less than 2 to the power of k plus 1 right but we know that this is a k plus b plus a right if you will open now a k plus b is less than this by assumption we can say that this is plus a right and now but now that's what I need actually so with sufficiently large beginning of this when we actually know that in the beginning my initial condition is already is already set I know that 2 to the power of k is greater than this and obviously since k is integer so it's obviously this thing is smaller than 2 to the power of k so a is smaller than 2 to the power of k so that's why I can say that this is less than 2 to the power of k plus 2 to the power of k because a is smaller than 2 to the power of k and that is and that is equal to 2 to the power of k plus 1 so our induction logic works fine so all I was just using that this thing is this thing is less than 2 to the power of k and that's why I substitute it I increased actually from a to the 2 power of k okay so that completes this so not only I have x divided by 3 to the power of x I actually have proven that any a x plus b divided by 3 to the power of x would also be infinitely small variable because because the numerator is less than 2 to the power of x eventually we are talking about x going to infinity so with sufficiently large x the numerator would be less than 2 to the power of x and 2 3rd to the power of x goes to 0 okay so any linear function how about quadratic function okay one more little exercise so we have x square divided by 3 to the power of x I would like to prove that this is also going to 0 and for this I would like to prove this in exactly the same fashion how to prove this well again n x n plus 1 so let me just prove that n plus 1 square would be less than 2 to the power of n with sufficiently large n how can I prove that well again let's do it by induction if n is equal to let's say what for instance 3 we have 4 square which is 16 no not enough n is equal to 4 this is 5 25 okay n is equal to 4 it's already good because this is 5 square 25 and this is oh no too small 5 this is 6 square 36 and 5 32 also not enough 6 7 square 49 2 to the power of 6 is what 64 okay now we have it greater alright so starting from n is equal to 6 I would like to prove so first n is equal to 6 n plus 1 is square is less than 2 to the power of n good now assuming that for n is equal to k I have k plus 1 square less than 2 to the power of k now let's set n is equal to k plus 1 so what do I have k plus 1 plus 1 so it's k plus 1 square plus 2 k plus 1 plus 1 now I know that this is less than this is less than 2 to the power of k plus 2 k plus 1 plus 1 now this thing is a linear function of k so basically I know that 2 to the power sorry 2k this is 2k plus 3 I know that 2k plus 3 less than 2 to the power of k with k sufficiently large so whatever the function but whatever the whatever the number which is sufficiently large to make this thing correct I will add to this one if it's necessary in this case actually I don't think it is necessary because this thing given for k is equal to 3 9 no 4 8 11 yeah with case equal to 4 it's already true and all greater so basically we started from 6 so we are safe so this thing we can always use so this thing is less than 2 to the power of k plus 2 to the power of k which is equal to which is equal to 2 to the power of k plus 1 and our induction is working fine so basically I can continue this type of proof if you wish for a polynomial of any degree and actually any polynomial p of x divided by 3 to the power of x would be eventually going to 0 because the polynomial itself will eventually be less than 2 to the power of power of x and that was my last problem which I wanted to discuss today how about you just go to the website Unisor.com and read again these problems whatever I have today discussed and try to solve them yourself that would be a very good exercise and basically that's it thank you very much and good luck