How do you apply the squeeze theorem to cos(x)/x as x approaches infinity?

[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.

 

[Hurkyl]

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.

 

[wais]

First of all, there is no need to feel embarrassed. Calculus is a challenging subject and it's completely normal to have trouble understanding certain concepts. The squeeze theorem can be a bit tricky to grasp at first, but with some practice and examples, you will soon have a better understanding of it.

To answer your question, the two functions that are "squeezing" the third one in this particular example are f(x) = cosx and g(x) = -1/x. These two functions are essentially bounding the third function, h(x) = (cosx)/x, from above and below. As x approaches infinity, f(x) and g(x) both approach 0, which means that h(x) must also approach 0.

You are correct in thinking that evaluating one of the functions at the limit is a key step in using the squeeze theorem. In this case, we can evaluate g(x) = -1/x at x = infinity, which gives us a limit of 0. Since h(x) is always between f(x) and g(x), it must also approach 0 as x approaches infinity.

So, in essence, yes, it is as simple as splitting up the function and evaluating one of the functions at the limit. However, it's important to remember that the other function must also approach the same limit in order for the squeeze theorem to be applicable.

I hope this helps clear up any confusion you may have had about the squeeze theorem. Keep practicing and seeking help when needed, and you will soon have a strong understanding of calculus concepts like this one. Good luck!