Originally Posted by

**fredrikhoeg** Hi, i got a question about the squeeze theorem.

Suppose |f(x)|<=g(x) for all x. What can you conclude about lim(x-->a) f(x) if lim (x-->a) g(x) = 0? What if lim (x-->a) g(x)=3?

Since $\displaystyle g(x)\xrightarrow [x\to a] {}0\Longrightarrow |g(x)|\xrightarrow [x\to a]{}0$ , we'd get that

$\displaystyle |f(x)|\leq g(x)\Longleftrightarrow -g(x)\leq f(x)\leq g(x)$ , and now apply the squeeze theorem to get the result for $\displaystyle \lim\limits_{x\to a}f(x)$ .

Can you see what doesn't work if the limits is 3?

Tonio

What i think i should do is to set it up this way:

-g(x)<=f(x)<g(x) to use the squeeze theorem. and i guess if g(x)=0, then -g(x)=0. But im not sure, and im not sure what happens if lim(x-->a) g(x)=3 with f(x)...

Thanks for help!