How does the squeeze theorem prove that a limit exists?

[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 don't understand is how it prooves this.

 

[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)?

 

[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?

 

[Faiza]

its also callled the sandwich theorem lol

 

[Galileo]

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?

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 $lim_{x<br /> \rightarrow 0}x^2\sin(1/x) =0$
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.