 Consider the limit as x approaches infinity of sine squared x over x squared plus one. Now, when you look at this one, you should have some concerns when you take limits as x approaches infinity or negative infinity of a trigonometric function. Because if you have something like sine, excuse me, take the limit of x approaches infinity of say sine of x, what happens? What should sine of infinity even be, right? There's really not a lot of sense of what that should be because the sine function just repeats itself over and over and over and over again. It will equally range from negative one to one. It just repeats itself periodically. So there really is no end behavior for sine of x because of the periodic nature of its graph. So how does that have a consequence here? We can't just plug in x equals infinity and expect anything to come from that. But what we can do, since we can't use the quotient limit property, is we actually can use the squeeze theorem to help us out here. So notice the following. If you take any quantity, any real number and you square it, that will always be greater than zero. So if you take f of x, who cares about f of x? If you square it, that's always gonna be great equal to zero. Square and something always makes it non-negative. And so sine squared will always be non-negative. We know that sine squared is gonna be greater than or equal to zero. But since sine also sits between one and negative one, if you square a sine, you're gonna see that sine squared will sit greater than or equal to zero like we saw, but also will be less than or equal to one, one being one squared as well. And so we can sandwich sine squared between one and zero. And then likewise, if you divide this expression by x squared plus one, well notice zero divided by x squared plus one will still be zero. You also get a one over x squared plus one, like so. x squared, like I said a moment ago, is non-negative. If you add one to it, it's definitely positive. And so when you divide these inequalities by positive, it won't change the direction whatsoever. So playing around with this, right? Sine squared over x squared plus one is gonna be greater than or equal to zero. Since sine is less than or equal to one, sine squared is less than or equal to one. And finally, if you divide everything by x squared plus one, you get these inequalities. This is the squeeze we want for the typical squeeze theorem. You'll notice that sine squared x over x squared plus one is sandwiched between zero and one over x squared plus one. We'll take the limit as x approaches infinity. Well, for the constant function zero, the limit will still be zero. So we see that this function approaches zero. But what about one over x squared plus one? If we add a little bit more detail there, it's a bottom-heavy rational function. You end up with one over infinity squared plus one. That's gonna become one over infinity, which is zero right here. So we see that the rational function also goes off toward zero. Well, since sine of x is sandwiched between two functions that go off toward zero, the squeeze theorem applies and tells us that the limit of sine squared x over x squared plus one will likewise equal zero. So we can still use the squeeze theorem even as x approaches infinity or x approaches negative infinity. This is particularly useful when you have to take limits as x approaches the extreme for periodic trigonometric functions because they themselves don't have a behavior at infinity, but in this case, the denominator dampers the whole thing down so that it does have a limit at infinity.