1. real hard problem

Suppose that $lim_{x\rightarrow a}$ $f(x)$ exist and is positive. Show that for any positive integer n we have
$
\lim_{x\rightarrow a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x\rightarrow a} {f(x)}}
$
.

2. Originally Posted by mancillaj3
Suppose that $lim_{x\rightarrow a}$ $f(x)$ exist and is positive. Show that for any positive integer n we have
$lim_{x\rightarrow a} \frac{n}{}\sqrt{f(x)}$ $= \frac{n}{}\sqrt{lim_{x\rightarrow a} {f(x)}}$ .
Those are meant to be n'th roots, right? $\lim_{x\rightarrow a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x\rightarrow a} {f(x)}}$ ?

The slick way to do this is to say that $\sqrt[n]{f(x)}$ is just the composition of the function f with the n'th root function $g(t) = t^{1/n}$. The function g(t) is continuous at all points t>0 (because in fact it is differentiable, with derivative $\tfrac1n t^{(1/n)-1}$). And the composition of two continuous functions is continuous. So the function $\sqrt[n]{f(x)}$ is continuous at $x=a$ (where $f(a)$ is defined to be $\lim_{x\to a}f(x)$).

3. thank you, is there another way to approach it?

4. Originally Posted by mancillaj3
is there another way to approach it?
You could use the identity $s^n-t^n = (s-t)(s^{n-1} + s^{n-2}t + \ldots + st^{n-2} + t^{n-1})$. This tells you that $|s-t| = \frac{|s^n-t^n|}{s^{n-1} + s^{n-2}t + \ldots + st^{n-2} + t^{n-1}} \leqslant \frac{|s^n-t^n|}{t^{n-1}}$ provided that s and t are positive.

Now suppose that $\lim_{x\to a}f(x) = \ell>0$. Put $s = \sqrt[n]{f(x)}$ and $t = \sqrt[n]\ell$ in the above inequality, to get $\bigl|\sqrt[n]{f(x)} - \sqrt[n]\ell\bigr| \leqslant \ell^{(1-n)/n}|f(x)-\ell|$. From that, you should be able to show that $\lim_{x\to a}\sqrt[n]{f(x)} = \sqrt[n]\ell$.