 Another way we have a finding limits is known as the squeeze theorem, sometimes also called the sandwich theorem. Actually, as it turns out, it's not particularly useful. It's only really ever used for one particular limit. But the squeeze theorem works in a very intuitive way. Suppose I have a balloon that enters a room, and maybe it bounces off the floor, ceiling, maybe it hovers a little bit, and it kind of bounces around inside the room for a little while. And here's the question you might want to wonder, what do you know about the height of the balloon? Well, intuitively, you know that if the balloon has to be someplace between the floor and the ceiling, you know that the balloon's height, wherever it is, has to be someplace between those two. And this is the basis for what's known as the squeeze theorem or the sandwich theorem. And suppose I have a function f, g, and h, and suppose I know that f of x, g of x, and h of x satisfy a particular system of inequalities. And this is going to be true for all x in some interval around x equals a, with the possible exception at x equals a. Note this accepting is not actually a typo. This is in fact a legitimate use of that particular spelling of accept. And so I'm going to have this inequality here. Well, first off, the limit as x approaches a, f of x, g of x, and h of x satisfies the same inequalities that the functions themselves do. Again, provided that all of the limits exist, all of the functions exist. And more importantly, if the limits of the outside functions, the f of x and the h of x, if the floor and ceiling functions have limits, and they have the same limit, then the middle function also has a limit, and that limit is going to be the common value. So here's an example. So suppose I want to find the limit as x goes to zero of x times sine of one over x. Actually, we don't need the squeeze theorem for this. We can actually reason our way to an answer as follows. Well, what do I know? Well, I know that as x gets close to zero, one over x is going to get close to plus or minus infinity, depending on whether x is going to be more than zero or less than zero. So one over x is going to plus or minus infinity. And the problem we have is that sine of one over x, well, this will be sine of a very large positive number, or the sine of a very large negative number. And to some extent, the sine of a very large or positive number is going to be unpredictable. And all we can really say about it is that it's going to be some number between minus one. What sine of one over x has no limit as x goes to zero. But that doesn't bother us because we don't care about sine of one over x, we care about x times sine of one over x. So whatever this is, it's going to be multiplied by a very small number. Remember, x is going to zero. So what do I have? Well, I have some number between minus one and one being multiplied by a very small number. And what I'm going to get there is going to be zero. It's going to be a small number. So I can conclude that this limit is equal to zero. And I can do that without applying any high-powered theorems. Well, I suppose I do want to apply some high-powered theorem. Amazon just sent me the squeeze theorem in the mail. I rip open the package, and I want to use it. And here's the problem. I'd limit as x goes to zero of x times sine of one over x. And I might go through the same analysis before. The limit as x goes to zero. Sine of one over x does not exist because as x goes to zero, one over x goes to plus or minus infinity. And the sine of a very large number is some number between negative one and one. And so I read the instructions carefully. And the squeeze theorem, if I want to use it, requires me to frame it between two functions. I want to find sine of one over x. I want it to be less than or equal to something and greater than or equal to something. And here's usually the problem in applying the squeeze theorem is finding those two framing functions. So what can I do? Well, there's my function. I want to think of something that is greater than or equal to x times sine of one over x. And one way that we might approach this, which works on this one particular problem, is that sine, because I know sine is always less than or equal to one, then x times sine of x is guaranteed to be less than or equal to x. In effect, I'm replacing sine of x with something I know that is larger than or equal to it, namely one. So I have x sine of one over x being less than or equal to x. By the same argument, I know that this is also greater than or equal to negative x because the least value of sine of one over x is negative one. And there, I've trapped my function in between these two. Now the art here is that because I'm eventually going to be taking a limit, I need to know that the right hand side and the left hand side not only have a limit, but that limit is the same number. And if I'm not careful in picking my two functions here, I'm not going to be able to evaluate that limit in the first place. So I check it out, c is x goes to zero, my right and left sides both go to zero, and that squeezes this function here in the middle. And so I can conclude that that limit is going to be zero. Again, it's worth keeping in mind that we don't have to go through all of this reasoning here if we're willing to go through this reasoning here. And the biggest problem with the squeeze theorem is that unless you choose your lower and upper functions carefully, they're not going to have the same limit. And that will leave us with no information about what the limit of the middle function is. So in general, the squeeze theorem is almost the last resort. Now that being said, there is one important limit where we do use the squeeze theorem, and that's finding the limit x goes to zero of sine of x over x. Now here, a little bit of analysis goes a long way. First of all, this is x going to zero. And the problem we have with this is that because the denominator is going to zero, then I might want to approach that zero from two different directions. So suppose I take a look at the limit is x goes to zero from above of sine x over x. And suppose that it's equal to some number. Well, one of the things I know from trigonometry is the sine of minus x is the same as minus sine of x, which means that the sine of minus x over minus x is equal to sine of x over x for all values of x. And what that means is that these two limits, the limit is x goes to zero from below, and the limit is x goes to zero from above, have to be the same, provided that the two limits exist. Because these values here, if x is a small negative number, are going to be the same as these values here for the same value of x, but positive instead of negative. So these two expressions are the same. And so the two limits have to be the same. Well, let's do a little bit of geometry. You can't do trigonometry without doing geometry. I suppose you could. It's just a thousand times harder. So let's think about the geometry of the trigonometric functions. And so I'll go back to my portion of the unit circle that's going to define my sine cosine tangent values. And let's consider this. I have this nice little blue triangle here. And I note that the area of that triangle is guaranteed to be less than the area of the sector. And so all I have to do is remember how to find the area of the triangle and the area of the sector. So the area of the triangle, well, this is going to be one half base is cosine. Height is sine. So there's the area of the triangle, one half sine cosine. And the area of the sector is determined by the length of the arc and the radius. It's one half r squared times the arc length. Well, this is the unit circle. So my radius is one. So the area of the sector, one half, one half, one squared times the arc length, which is going to be our x value. So there's the area of the triangle. There's the area of the sector, x over two. And through some very, very, very complicated algebra, I get sine of x over x less than or equal to one over cosine x. Well, how about the area of the sector? Well, the area of the sector is going to be less than the area of this triangle. And I have a larger triangle there. And let's see. So again, we can calculate the area of the sector. That's going to be x over two once again. And the triangle, one half base times height. So our base is this is the radius of the circle. That's one. The height, well, this is the tangent of x. And so the area of our triangle, one half tangent x. So x over two is less than or equal to one half tangent x. Remembering that tangent is sine of our cosine and again doing some very, very, very, very complicated algebra. We can go from this expression to cosine x less than or equal to sine of x over x. And I've trapped my function in between two other functions. And we don't know that this will do us any good. But if the outer limits have limits as x approaches zero and they're the same value, then I can find the inner values. And it turns out that they do. So I'll take the limit as x goes to zero from above of cosine and one over cosine, those are both equal to one. And that says that the limit of x approaching zero of sine x over x also equal to one. And because the limit from above is the same as the limit from below, then our limit no plus, no minus is equal to one.