# Aqueeze

• Jul 13th 2008, 01:18 PM
HELLMACH
Aqueeze
I dont understand the idea and proof behind Squeeze Theorem
• Jul 13th 2008, 01:46 PM
janvdl
Quote:

Originally Posted by HELLMACH
I dont understand the idea and proof behind Squeeze Theorem

Hmm, now let me recall the lecture I had myself :D

Lets say we want to know what the limit $\displaystyle \lim_{x \rightarrow 0}$ approaches for the function $\displaystyle x \ \sin(x)$

Now we know $\displaystyle \sin(x)$ oscillates between $\displaystyle 1$ and $\displaystyle -1$

So we can say:

$\displaystyle -1 < \sin(x) < 1$ (We know the value of the limit must be somewhere between these two values.)

But the original question asks for $\displaystyle \lim_{x \rightarrow 0} x \ \sin(x)$

So multiply throughout the equation with $\displaystyle x$

$\displaystyle -x < x \ \sin(x) < x$

We are given that $\displaystyle x$ approaches $\displaystyle 0$

$\displaystyle 0\leq \lim_{x\to0}( x \sin x )\leq 0$

Now observe, the value must be 0. It lies between 0 and... well uhm... 0 :D So it must be 0 itself.

And so we have calculated the limit! (Clapping)

(The proof behind it will be in your textbook, go through it again now that I've shown you the idea behind the theorem.)
• Jul 13th 2008, 02:07 PM
flyingsquirrel
Hi
Quote:

Originally Posted by janvdl
We are given that $\displaystyle x$ approaches $\displaystyle 0$

$\displaystyle 0 < x \ \sin(x) < 0$

I think it'd make more sense to write $\displaystyle 0\leq \lim_{x\to0}( x \sin x )\leq 0$.
• Jul 13th 2008, 02:10 PM
janvdl
Quote:

Originally Posted by flyingsquirrel
Hi

I think it'd make more sense to write $\displaystyle 0\leq \lim_{x\to0}( x \sin x )\leq 0$.

Tonight is not my night for doing math...(Worried)

Thank You. (Nod)