# Squeeze theorem

1. Oct 31, 2004

### kdinser

Ok, this is kind of embarrassing. I'm halfway through calc II and while the squeeze theorem makes sense to me when I read it, I don't see how it's applied. I feel like I must be missing something very fundamental for this to not make sense.

$$\lim_{x \rightarrow \infty}(cosx)/x = 0$$

What are the 2 functions that are squeezing the 3rd? Once you know the 2 functions do just evaluate one of them at the limit?

Is it as simple as just splitting up the function? $$(1/x)(cos x)$$ and then evaluating $$1/x$$ at infinity. That would certainly yield 0.

2. Oct 31, 2004

### Hurkyl

Staff Emeritus
Usually, you use the squeeze theorem by identifing a piece of your function that you can bound.

Let me do an example in a slightly silly way:

$$\lim_{x \rightarrow \infty} \frac{x^2 + \ln x}{x^2 + 1}$$

Now, when x is large, we have the bound $0 < \ln x < x$. With a little work, we can prove:

$$\frac{x^2}{x^2+1} < \frac{x^2 + \ln x}{x^2 + 1} < \frac{x^2 + x}{x^2+1}$$

for large x. (such as x > 1)

We can evaluate the limits on the left and right hand sides: they both come out to 1. So, the original limit must also be 1!

That's how the squeeze theorem is usually used: you identify a piece of the function that you can bound, then substitute in the bounds to get an inequality to which you can apply the squeeze theorem.