limits using the squeeze theorem?

Question

I need to find the limit as x--> infinity of cosx/3x, When I look at the graph I see that the function is obviously periodic and oscilates between higher and higher values of y the further you get from the origin in either dirrection. How can I use the squeeze theorem to find the limit of cosx/3x as x goes to infinity?

Answer 1

I do not think that this problem requires squeeze theorem. Think of it this way: as x goes to infinity, 3x will go to infinity and cos x will oscillate between -1 and 1. So as the denominator gets huge, and the numerator stays really small, the limit as x goes to infinity is 0.

Answer 2

Squeeze Theorem

Answer 3

Old Paul: That's exactly the sort of situation the squeeze theorem is for.

Nicholas: You can still have limits as x tends to infinity; that is a perfectly well-defined and precise concept.

To answer the question, you find two functions which are upper and lower bounds for your function and have easily computed limits (that are the same!). Here, the obvious functions to try are -1/3x and 1/3x. For any x > 0 we have
-1 / 3x ≤ cos x / 3x ≤ 1 / 3x.
Furthermore, lim (x->∞) -1 / 3x = lim (x->∞) 1 / 3x = 0.
So by the squeeze theorem we have
lim (x->∞) cos x / 3x = 0.

Answer 4

You can't, the squeeze theorem does not apply. It requires two functions to approach the same limit at a point (which your function will be "squeezed" between). Infinity is not in the domain of your function. For that matter, its not even a number, so you cannot have a point at infinity.

But it sounds like you typed your equation wrong. (cos x) / 3x has a horizontal asymptote at x = 0, it doesn't diverge like you describe.

The below answer is a correct explanation of how to analytically find the limit, but is inaccurate in one part of the process. The function is not squeezed between 0 and 0, because the function crosses x = 0. The below answer does not use the squeeze theorem as I am familiar with it.

Answer 5

limit as x--> infinity of cosx/3x

use the fact that

-1 ≤ cos(x) ≤ 1

divide by 3x all terms of the inequality (x not zero)
-1/3x ≤ cos(x) /3x ≤ 1/3x

as x ---> infinity
1/3x ----> 0
cos(x) /3x is squeezed between 0 and 0, it has to be zero (cos(x) is bounded when x ---> infinity)