Using Squeeze Theorem To Solve An Obvious Problem

Homework Statement

So we have to evaluate this using the squeeze theorem:
http://www4c.wolframalpha.com/Calculate/MSP/MSP230197da409360513h500003f12d5h334he7b32?MSPStoreType=image/gif&s=56 [Broken]

The Attempt at a Solution

Well it's pretty obvious that it will go to zero. Reason being that anything divided by infinity is zero...but how do you evaluate this with the squeeze theorem? There's only one equation given, and it's easily solved.

Thanks in advance.

Last edited by a moderator:

Answers and Replies

Landau
Science Advisor
Well it's pretty obvious that it will go to zero. Reason being that anything divided by infinity is zero...
That is a pretty lousy (or extremely sloppy formulated) reason :tongue:
You seem to think that $$\lim_{x\to \infty}\frac{f(x)}{x}=0$$ no matter what f is. That is certainly not true (take e.g. f(x)=x, or f(x)=x^2).

The key observation here is that the sine is bounded, in particular $$|\sin x|\leq 1$$ for all $$x$$, in other words $$-1\leq \sin x\leq 1$$. Try to use this.

Hmm, the image of the function vanished.

So you're saying I should use -1 and 1 as my other two functions that go to infinity to evaluate sine at infinity?

Since sinx only goes between 1 and -1, it will never make sinx/x go higher or lower than 1/x or -1/x, respectively. Can you understand why?

I guess. Because trig functions have a range equal to that of +-their amplitude?

The amplitude of sinx is 1. Will numbers between 0 and 1 (ignoring the negative values of sinx) multiplied with 1/x ever make 1/x be greater than it normally is?
After than, consider the negative values of sinx and -1/x.

Landau
Science Advisor
Hmm, the image of the function vanished.
I don't quite understand what you mean by this.
So you're saying I should use -1 and 1 as my other two functions that go to infinity to evaluate sine at infinity?
No, you don't need to 'evaluate sine at infinity' (this not a meaningful expression).

Could you please state the squeezing theorem? Then you will immediately see what to do with the estimate of sine being between -1 and 1.

I just got it, thanks for the idea bouncing guys. It should be evaluated as "-1/x <= sin(x) <= 1/x".

Yay!