# Squeeze Law (help)

1. Oct 10, 2004

### thennigar

need someone to explain this law to me. i understand the fact that if a function exists between two other functions on a graph then it can be squeezed hence the "squeeze". What i dont understand is how it prooves this.

2. Oct 10, 2004

### Galileo

The squeeze theorem I`m familiar with says that if:
$$\lim_{x \rightarrow a}f(x)=b$$
$$\lim_{x \rightarrow a}g(x)=b$$
and $f(x)>h(x)>g(x)$ for all x in a certain neighbourhood of a.
Then
$$\lim_{x \rightarrow a}h(x)=b$$

The proof follows easily from the definition of a limit.
If $|g(x)-g(a)|<\epsilon$ and $|f(x)-f(a)|<\epsilon$,
what does that mean for h(x)?

3. Oct 10, 2004

### thennigar

ok. that makes sense. now how can you apply that to say g(x)=(xsin(1/x)) x cannot = 0.

would abs(x) be your two functions that squeezes g(x)? then if the limits of
-x and +x are the same then the limit g(x) must also be this?

and one last question, how do you know when to use this law?

4. Oct 10, 2004

### Faiza

its also callled the sandwich theorem lol

5. Oct 11, 2004

### Galileo

Not sure what you mean exactly, but I think you have the right idea.
Anyway, don't use $\pm|x|$, since it doesn't bound your function.
Suppose you want to show that $lim_{x \rightarrow 0}x^2\sin(1/x) =0$
Because $-1\leq sin(1/x) \leq 1$, we have $-x^2\leq x^2sin(1/x) \leq x^2$.
Then apply the squeeze theorem.

There are no rules for when to use it. But it's commonly used when there's an oscillating and bounded term like here.

Last edited: Oct 11, 2004