 Hi, I'm Zor. Welcome to Unizor education. I would like to continue doing some exercises with function limits and today we will do it with the argument which goes to negative infinity. Now this lecture is part of the course of advanced mathematics for high school students and teenagers presented on Unizor.com. I recommend you to view it from this website because the website has explanation written notes for each lecture plus there is a functionality like passing exams, enrolling, etc. and site is completely free, so all right. Now, talking about going to negative infinity, well, maybe for some purely aesthetically reasons, you would prefer to go to positive infinity as a little bit more familiar, if you wish, way of changing argument. Now, how can you approach the problems with negative infinity? Here is what I suggest. We all know about compounded functions and the limit of the compounded functions. Let me just remind you if you have function f at x, which goes to some limit l as x goes to some limit point r and another function which goes to m as its argument goes to l pay attention, this is l and this is l then you can conclude that the combination of these functions, the compounded function, will go to m as x goes to r, this one. As x goes to r, function f at x goes to l, so the argument will go to l and if argument goes to l, then the function goes to m. Now, this obviously is true not only for real numbers, r and l and m, but also for infinities. So what I'm suggesting right now is the following. I will use minus infinity here and I will use a function minus x. So what happens with this function minus x is if x goes to minus infinity, it's a negative infinity. Well, obviously it goes to positive infinity, right? Now, so l is equal to positive infinity now, right? And now let's go to a concrete example. So r is negative infinity. Okay, my examples are exactly the same as in the previous lecture and that what makes it a little bit easier. So the first one is 2x squared plus 3x plus 4 divided by 3x squared plus 4x plus 5 as x goes to negative infinity. Well, in the previous lecture we were talking about positive infinity, right? All right. Fine, so what do we do? Well, let's use the combination function. The function which would be so if this is f, this is g of x and I know by the way that if x goes to plus infinity, my function goes to 2 thirds, right? Because the powers are the same x square and x square. Now, what if I will do this? According to this theorem, where f at x is this and x goes to minus infinity, right? It's supposed to go to the same limit. Now, how can I obtain this function in such a way that if I will substitute minus x, it will go, it will present the function which I really need. Here it is. Instead of function g, let's consider function g1 which is 2x square minus 3x plus 4 divided by 3x square plus minus 4x plus 5. Now, if I substitute instead of x, I substitute minus x, then this function would be exactly like this one. Now, what do I know about limit of this function? Well, if you remember since my powers are the same, limit would be the ratio of the coefficient. So it will go to 2 thirds, the same 2 thirds as this one, by the way, which happens to be. So now, since I know where this function will go, I can substitute g of g1 of f at x and that would be my function g of x, right? Which is 2x square plus 3x, etc. And I know how this function behaves and it will go to 2 thirds. So I expressed my initial function using the compound function, behavior of which components I know. Now, on another hand, I obviously could use exactly the same principle for this particular limit as I did for plus infinity. Namely, I can say that this is equal to 2 thirds times x square times 1 plus 3 divided by 2x, right? Plus 4 divided by 2x square divided by x square again. 1 plus 4 divided by 3x plus 5 divided by 3x square. And obviously, with x going to minus infinity, our denominator on these four cases will be infinite by absolute value, which means my fractions will be infinitesimals and the whole thing would actually be 1 divided by 1, which is 1 and the limit would be 2 thirds. So these are very simple cases when you really can apply your compounded theorem or you can just do it directly as easy. Now, let me consider a couple of other examples which are very much similar. Now, the first one has the third power here. Now, what happens in this case? Well, if you do it directly like I just did before, you would have something like 2x divided by 3, and then in parentheses you have x square and x square, obviously, which will cancel each other. Now I will have 1 plus 3 divided by x square plus 4 divided by actually 2x square, 2x cube. And here I will have 1 plus 4 divided by 3x plus 5 divided by 3x square. And the same thing here. These are all infinitesimals. These x squares are canceling, and I have this x which doesn't cancel anything. These are constants, so everything basically is determined by x. And the limit obviously is if x goes to negative infinity, the whole thing will go to negative infinity. And finally, what if I have higher power in the denominator? Well, in this case it would be 2x square 3x times x square on something which is really a constant, well, not a constant, but constant plus infinitesimal value. And obviously everything would be determined by this one. And since x goes to negative infinity and this is denominator, obviously the fraction will go to zero, but from the negative side. All right, next. Next example is my function g of x is equal to x times sin 1 over x. Now in the previous lecture, when x was actually going to plus infinity, we basically concluded that this limit would be equal to 1. Okay. Now, how can I obtain this function using the compound things? Well, how about this? I know that sin is an odd function, right? So sin of minus something is equal to minus sin. So what I will do is the following. I will have function which is minus x times minus sin 1 over x. Let's say this is g1 of x. Now, if I substitute instead of minus x, I substitute x, this will go to x. And this, this will go to minus x, but sin is an odd function. And with this it will be minus again. So it will be minus and minus it will be plus. So this function is actually a good candidate because g1 of f or x is equal exactly my gx, g of x, where f at x is this one, right? So what is the limit of this function? Well, look, this is minus and minus, so it's the same as plus. So it's the same as the one which we had before. So basically g1 and g are exactly the same function. G1 and g are exactly the same functions. And that's why the limit is exactly the same as we go to minus infinity. And on the other hand, you can always judge it slightly differently. You can always have it as sin 1 over x divided by 1 over x. If x goes to negative infinity, this goes to 0 and this goes to 0. And we basically know that sin x divided by x, if x goes to 0, really is approaching 1. So that's why all these examples are the same basically and the limit is 1. Now one more example. I have to find limit this. As x goes to minus infinity. Now if this is my g of x, let's consider a different function. g1 of x is equal to x divided by 3 to the power of x. This is the function which we have considered before. Now, if I substitute minus x instead of x, what will be minus x divided by 3 to the minus x? Now negative power, as you know, is inverse. So it's equal to minus x times 3 to the power of x. Well, almost the same as we need, right? So to make it better, let's do it this way, okay? So if we consider this function, then g1 of f of x is exactly our g of x. So the limit of this thing is basically the same as the limit of this thing, right? Now, what is this limit as x goes to plus infinity? Well, we already did it in the previous lecture and the limit was 0, but it was a plus sign, but it doesn't really matter because it's 0, right? So since this limit is 0, then the limit of this thing as x goes to negative infinity is the same 0. So 3 to the power of x, where x is going to the negative infinity, let me just remind you the graph of this function. This is my 3 to the power of x and this is my x. When I multiply it, you see this is growing in absolute value to a negative infinity, right? But this is going to 0, so it's infinity times 0, it's indeterminate, right? However, this thing is much quicker goes to 0 than this one goes to infinity and that's why the result of their product tends to 0, okay? But from the negative side, because you're basically multiplying something negative, which is x, by something positive, which is 3 to the power of x. So eventually, if you will multiply these two curves, now in this case it's 0, right? Then it's probably growing a little bit and then it goes to 0. Now here, obviously the multiplication would be something like this. So the curve would be like this. And the next one, which is very similar, which is x square. Now how should I do this? Same thing. I consider this function as my g1 of x. And I know that as x goes to plus infinity, this thing goes to 0, right? Now g1 of f of x is exactly what my g of x is, where f of x is minus x. So the limit is exactly the same and it will be reached when x goes to minus infinity. And again, that's because power grows slower than exponent. And whenever we are inversing this situation, whenever we are multiplying by 3 to the power of minus x, that's actually exactly the same. 3 to the power of minus x as x goes to negative infinity goes to 0 faster than x square increasing in its value. So the graphs would be like this. This is x square and this is my 3 to the power of x. Yeah, 3 to the power of x, right? And if you multiply them as x goes to minus infinity, this thing is faster going to 0 than this one to infinity and the result would be 0. So it's something like this, something like this. All right. So these are examples of certain limits related to negative infinity. And as I explained, in some cases you might actually directly calculate these limits. In some cases you might resort to theorem about compounded functions if it makes your life easier whatever it is. And I would suggest you to try to solve these problems again just by yourself. I think it would be a very good exercise. All right, that's it. Thanks very much and good luck.