Sandwich theorem?

[EriLynn.]

Okay so i've been reading my text book and it talks about using the sandwich theorem to compute questions like "limit as x approaches infinity of (sinx/x)" and im not sure of how to used this sandwich theorem because all the book says it the following:
" If f(x) is less than or equal to g(x) which is less than or equal to h(x) for all x in an open interval that contains c (except possibly at c) and the limit as x approaches c of f(x) is equal to the limit as x approaches h(s) equals L then the limit as x approaches c of g(x) equal L..."

Can someone please explain how i use this theorem for the above question, or what kind of steps i need to follow to show this theorem??
Thanks.

 

[galactus]

This example is in most calc books. If you google it you can find it. It is one of the most common examples when demonstrating the Squeeze/Sandwich Theorem

It is based on a unit circle sector. Just google 'squeeze theorem' and I am sure you'll find it.

EDIT: Oh, you said as x --> infinity, and not 0. OK. Let's do it this way then:

\(\displaystyle \lim_{x\to \infty}\frac{sin(x)}{x}\)

Sine ranges from -1 to 1:

\(\displaystyle -1\leq sin(x)\leq 1\)

Divide by x:

So, \(\displaystyle \frac{-1}{x}\leq \frac{sin(x)}{x}\leq \frac{1}{x}, \;\ x>0\)

But \(\displaystyle \frac{-1}{x} \;\ \text{and} \;\ \frac{1}{x}\to 0 \;\ \text{as} \;\ x\to {\infty}\)

Therefore, \(\displaystyle \lim_{x\to \infty}\frac{sin(x)}{x}=0\)

See?. Since sin(x)/x is Sandwiched between -1/x and 1/x, and since both -1/x and 1/x head toward 0 as x gets larger and larger(limit is unbounded)*, then sin(x)/x must have limit 0 as well.

*I do not like to say "as x approaches infinity". Nothing can approach infinity, though we know what is meant. To say "x approaches infinity" is an 'sloppy' way of putting it. We could say that x increases without bound. See what I mean?.