# Limits using the squeeze theorem

Hey I am doing some revision and I came across this problem. I'm sure it uses the squeeze theorem however I have never seen it used for a function like this. It just seems like really different compared to most of the problems I have used the squeeze theorem for.

$$x^4\leq f(x)\leq x^2 \quad \text{ for } x \in [-1,1]$$

and

$$x^2 \leq f(x) \leq x^4 \quad \text{ for } x > 1, x < -1$$

1) At what point $a$ does $\lim\limits_{x \to a}f(x)$ exist?

2) For each of these values determine the $\lim\limits_{x\to a} f(x)$

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Your first question doesn't make any sense. Please explain further. –  Fly by Night Aug 22 '13 at 13:30
I added some extra info @FlybyNight –  Ghozt Aug 22 '13 at 13:34

Clearly, if $|x| > 1$, we have

$$x^{2} \leq f(x) \leq x^{4}$$

and since $x^{2}$ and $x^{4}$ both go to infinity, so does $f(x)$.

If $|x| \leq 1$, then

$$x^{4} \leq f(x) \leq x^{2}$$

If $|x| = 1$, then $x^{2} = x^{4} = 1$ and hence $f(x) = 1$.

If $|x| < 1$, then $x^{4}$ and $x^{2}$ both go to zero and hence so does $f(x)$.

Hence, to summarize,

$$\lim\limits_{|x|\to \infty}f(x) = \infty$$

$$\lim\limits_{|x| \to 0} f(x) = 0$$

$$\lim\limits_{|x| \to 1}f(x) = 1$$

For other points, the information given is not sufficient to derive if $f$ has a limit.

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It really makes sense now, thank you. –  Ghozt Aug 22 '13 at 13:48