# Squeeze Theorem

1. Sep 18, 2007

### physstudent1

Hey could someone explain the squeeze theorem to me a little; I understand you want 2 equations that "squeeze" another one into between them sothat you can find they're limits and find the equations limit but how do you find the 2 equations that squeeze the original one in?

2. Sep 19, 2007

### Pseudo Statistic

There's no general way of finding two functions that satisfy the criteria...
You just want to find two functions $$g$$ and $$h$$ such that:
$$lim_{x\rightarrow a}g(x)=lim_{x\rightarrow a}h(x)$$ and
$$g(x)\leq f(x) \leq h(x)$$ for all $$x$$ within ssome neighbourhood of $$a$$. Then, the squeeze theorem tells you that the limit those two have at $$a$$ is the same as the limit $$f$$ has at $$a$$.
Like I said-- the choice of your $$g$$ and $$h$$ is completely arbitrary-- you just want them to satisfy those conditions.
Sometimes, e.g. when you have $$\frac{1}{x}sin(x)$$, boundedness helps. Observe:
$$-1 \leq sin(x) \leq 1$$ (Property of the sine function)
This implies $$-\frac{1}{x} \leq \frac{1}{x} sin(x) \leq \frac{1}{x}$$. (Whenever $$x>0$$)
As you can see, the limit as $$x\rightarrow \infty$$ of the left and right hand sides of the inequality match and equal 0, and so the limit of the function $$\frac{1}{x} sin(x)$$ is 0. This is a classic application of the squeeze theorem.
I hope that helped.

3. Sep 26, 2007

### CrazyCalcGirl

I get that part of it, but my question is on the 2nd part of the squeeze theorem.

It states abs(g(x))<=M for all x not zero.
then lim f(x) x g(x)=0 as X--> a. Can someone show me the proof to this part and explain it. The book shows no further information and I'm confused as to what it means.

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