1. ## The Squeeze Theroem

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?

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!

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

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

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!
.

3. Thanks, so I understand that since lim(x-->a) g(x)=0, the lim(x-->a) f(x) = 0, since lim(x-->a) -g(x) is also 0.

hmm, what doesnt work when lim(x--a) g(x)=3? maybe that lim(x-->a) -g(x) isnt the same anymore?

4. Yes, $\lim_{x\to 3} g(x) \ne \lim_{x\to 3} -g(x)$ so $f(x)$ isn't being "squeezed" at $x=3$.