[MUSIC] Now let us consider some examples of the creation of functions limits basic definition. So let's start with the idea of the limit of, for example, function x squared in some arbitrary point a. So what we are going to do, we are going to compute the limit of x squared if x approaches a, right? So firstly, one can understand that this limit is actually easily calculated by arithmetic rules that we are going to cover further in our lectures. But for the sake of the example, let us just try to find it by the definition. So as we established, there is two basic definition of the limit of the function. One concerning deviations on the epsilon-delta language, and one concerning sequences, all right, which is more useful for proving absence of the limit of the function. So as one could imagine, the limit of x squared if x approaches a is a squared, it's pretty simple. Let's try to prove it by our first epsilon-delta definition. In order to do so, first let us just revisit it by writing it down. For any positive deviation, you can go as close to the point where the limit is taking, but not exactly at this point. So the definition between function and its limit, I'm going to use capital A is smaller than established threshold. Okay, that was very imminent. Let us substitute what we actually have here. We know what is our function, it's x squared, right? And what is our limit here? Limit is a squared. So we are looking at the absolute value of x squared minus a squared. Let us rewrite it here. Okay, so what we are going to do, first of all, I'm going to do some school trick. Since we are looking at the difference of the squares, we will factor it into two brackets, (x- a) multiplied by (x + a), so that's easy. More importantly, the absolute value of the multiplication is multiplication of absolute values. Thus we get that we are looking at absolute value of (x- a) multiplied by (x + a). The first part here, the first multiplier here is pretty nice because it is, by the definition of the limit, supposed to be bounded by 0 from the lower side and delta, which we're looking for, from the upper side, right? But what one should do with the second multiplier, since it is dependent from the x. Well, let us just consider that our deviation from the point a does not exceed, well, the point a itself. Or in other words, the absolute value of x plus a do not exceed two a's. So we're going to just use another inequality here. We are going to consider that the first multiplier is no more than delta, which we are looking for, right? That's how we. And the second multiplier here is not greater than 2 absolute values of a. We are going to write absolute value of a here, just because we can assume that the limit is also can be taken in negative point here for negative a's. So what we got here, we got here that basically our deviation do not exceed delta multiplied by 2 absolute value of a's. And if we found a delta in which case there's expression delta multiplied by 2 absolute value of a is smaller than epsilon, then we actually get our desired inequations that the deviation does not exceed the threshold, right?. So in order to find that, we just need to write that this where delta is actually epsilon divided by 2 absolute value of a, right? Well, of course, one could argue that this results into the equality here, right? So this sign is kind of doubtful. But since the previous sign is pretty strict inequality, so we are going to say that this very relation still holds. Or you can just substitute our delta with half of a delta, if you really like it. But that still holds. Basically we just proven that our limit of x squared is a squared at arbitrary point a. So for the second example here, let us just assume that we are looking at some function which is clearly problematic at the limit point. In order to do it, we are going to, Consider the limit of the sine of 1 divided by x if x approaches 0. And I'm going to write question mark as our answer. Okay, so we need to answer two basic things here. Does it have a limit? And what its value if it does? So well, if one just try to draw a sketch of this graph, what we are going to have. We are going to have, well, first, two axes, and then we're going to have function which is oscillating. And since we are getting as close, close to 0, thus 1 divided by x is getting closer, closer to infinity, right? Thus basically it moves from 0 to 2 p, which is kind of the spirit of the sine function, faster and faster, right? So what we should draw here, we should draw here function that starts to oscillate faster, faster, faster, and faster. Okay, basically that's a nightmare, right? So does it have a limit or not? We are going to try to prove that it does not by using our second idea. That for any approach, for any sequence of arguments that approach our limit point, the limit of the values of the functions will approach all the same value, which is a limit of the function at this very point, right? So in order to do, it we're going to to imagine basically two sequences. First one I'm going to call xm, and this will be the sequence where sine, for example, equals to 0. And in order to do it, we are going to say that argument with high function should be, for example, 2 multiplied by pi multiplied by n. If we rewrite this in more direct form, xn is 1 divided 2 pi n, which, let us check, approaches 0 if n approaches Infinity. Thus it's valid sequence to approach our limit point, our x approaches 0, right, this one. So that's our first one. And then we are going to consider some arzo sequence of arguments, I'm going to call it yn. In which our function actually has, for example, our sine function has value 1, which is half a pi plus 2 multiplied by pi multiplied by n. Once again, in direct form, it is 1 divided by half a pi plus 2 pi n. And it approaches 0 since we get a constant divided by the function, which is it approaches plus infinity. Okay, that's fine. So basically we have two sequences. And then let us just understand that the sine of 1 divided by xn equals to 0, thus approaches 0. And sine of 1 divided by yn equals to 1, thus approaches 1. And this is not the same limits, thus the function doesn't have a limit, there is an absence of limit. So we are going to write that this limit do not exist. [MUSIC]
