 In this final video for lecture 15, as we've been discussing, like in the previous lecture as well, about vertical and horizontal asymptotes of functions, in particular, we're interested in the in behavior of various functions. There's one other family of functions that we run across in calculus that I do want to talk about that has a horizontal asymptote, and that's going to be the arctangent function. You know, we've kind of ignored the inverse trigonometric functions a lot in our lecture series, but arctangent's a pretty good function in its own right. It doesn't have the restricted domains like arc sign and arc cosine have, and it does have this asymptotic behavior that you see the graph of arctangent given right here. As x goes towards infinity, we're going to see that tangent inverse of x will approach pi halves, and as x goes towards negative infinity, we're going to see that arctangent will approach negative pi halves. And this comes from the fact that arctangent is the inverse function to tangent. Since tangent has vertical asymptotes at pi halves and negative pi halves as x ranges from negative pi halves to pi halves, we see that the inverse function, which will switch the roles of the x and y coordinates, we see that those vertical asymptotes turn into horizontal asymptotes, and then the range will range from negative pi halves to pi halves. And so for the tangent function, its range was all real numbers. Arctangent, its domain is now all real numbers. The roles of x and y get swapped around. And so those will come up as we compute limits as x approaches infinity right here. So consider the limit as x approaches infinity of sine of arctangent of x. Well, because sine is a continuous function, you can bring it out of the calculation here because continuous functions can be brought out of limits. So we're going to get sine of the limit of arctangent here of x as x approaches infinity, like so. And so as x approaches infinity here, arctangent's going to approach pi halves. It'll approach pi halves from below. So this will become here sine of pi halves, specifically or approaching it from the left side. What that really means is we're computing the limit as x approaches pi halves from the left of sine of x. That's really what we're doing. That's the more precise thing here. But this, because of the continuity of sine, we just have to compute sine of pi halves, which we know sine pi halves, that's 90 degrees. Sine is going to equal one in that situation. So knowing the horizontal acetote of arctangent is going to be very, very useful. Now, I do want to approach this question in a slightly different perspective, because whenever you deal with inverse trigonometric functions, you can associate those as an angle, right? So we say theta is equal to arctangent of x, which actually tells us that tangent of theta is equal to x, or in other words, x over one, for which associated to this angle, we can connect a right triangle, a right triangle whose angle in question is exactly this angle theta right here, for which if the tangent ratio, opposite over adjacent is x over one, we can fill those in as x over one, the opposite side's x, the adjacent side is one, and then by the Pythagorean equation, the other side, the hypotenuse will be the square root of one plus x squared. Using this, notice what we're trying to compute here? We're trying to consider what is sine of theta of arctangent of x is theta, then we try to compute what sine of theta is. So sine, which is opposite over hypotenuse, becomes x over the square root of one plus x squared. We could substitute sine of tangent inverse of x with the function x divided by the square root of one plus x squared. Now this function right here resembles types of infinite limits we were considering previously. Notice if we take the limit as x approaches infinity here of x over the square root of one plus x squared. Well, if we look at just the dominant terms, the dominant term on top is x the only term, the term on the bottom that's dominant will be the square root of x squared, right here is x approaches infinity, for which the square root of x squared is actually the same thing as the absolute value of x, x divided by the absolute value of x, but as we're approaching positive infinity, approaching positive infinity implies that x will eventually be positive, and as such the absolute value is no different than x itself. And so this will simplify just to be the one that we saw previously. So this gives us a second way of computing that same problem if we wanted to compute it using sort of like this algebraic approach as opposed to the trigonometric approach we started with. Either approach is perfectly fine. I thought the yellow one was fairly straightforward, so that's the one I would recommend if we needed to. Let's consider the limit as x approaches negative infinity of arc tangent of e to negative x. Kind of like we saw with the previous example, since arc tangent is actually a continuous function, we can bring it out of the limit calculation. So we're going to get tangent inverse of the limit of e to the negative x as x approaches negative infinity right here, in which case then we essentially we have to compute tangent inverse of e to the negative negative infinity. This is really going to be e to the infinity. That's going to be arc tangent of infinity. What does that mean? Now when people, some people don't like this arithmetic with infinity here. It's like, oh, because it's dangerous. It's dark magic. Oh, you're going to, you know, curse us all. Well, when we say things like arc tangent of infinity, this is really just an abbreviation for limit notation. We're saying take the limit as x approaches infinity of arc tangent of x. That's really what we're asking. And so arc tangent of infinity is asking, what's the horizontal asymptote of arc tangent on the right? What is the in behavior on the right hand side? And as we observe on the previous slide, that's going to be pi halves. So we see that the limit as x approaches negative infinity of arc tangent of e to the negative x is going to be pi halves. One last example. Let's consider the limit as x approaches two from the left of arc tangent of one over x minus two. Much like we did on the previous one, I'm going to pull the continuous function out of the limit. So we're going to look at arc tangent of the limit of one over x minus two as x approaches two from the left. For which case we can plug in that two from the left, we're going to take arc tangent. We're going to get one over two from the left minus two. So as we're considering, we're approaching two, but taking numbers that are less than two. So if you take a number less than two and subtract from a two, we're going to get a number that's really close, but less than zero. So we end up with one over zero minus as we've seen previously, one over zero minus as we take if we take negative numbers that get really, really small in terms of absolute value, that's going to explode towards negative infinity. So we're getting arc tangent of negative infinity right here. And like I said, previously on the last example, if you really get like your, I don't know your, your, your buttons in a twist or something, I don't know, when it comes to doing arithmetic with infinity or negative infinity, remember that these symbols of using infinity is always shorthand for a limit. We're taking the limit as x approaches negative infinity of arc tangent of x. That's all we're trying to do here. So don't, don't worry about it. It's just an abbreviation of such a thing. For which case the, the horizontal asymptote on the left of arc tangent is going to be negative pi over two.