Prove $\lim_{x \to p} \sqrt[n]{f(x)} = \sqrt[n]{L}$, where $\lim_{x \to p} f(x) = L$. $f(x), L \ge 0$ if $n$ is even. $f(x)$ is a real valued function.
This seems to be a standard result, but I failed to prove it. I basically don't know what I can do with $\sqrt[n]{a} - \sqrt[n]{b}$.
We might prove this by the property of its inverse function $f(x) = x^n$? I'm not sure about what theorems I can use though. I would like to hear how this way can work out, although I prefer direct argument on the $n$th root function.
HINT:
To show that $x^{1/n}$ is continuous, use the identity
$$a^{1/n}-b^{1/n}=\frac{a-b}{a^{1-1/n}+b^{1/n}a^{1-2/n}+\cdots +a^{1/n} b^{1-2/n}+b^{1-1/n}}$$