How do I prove, using the definition, that the nth root is a continuous function? Stuff I've tried:
\begin{align}
|x^{\frac 1 n}-a^{\frac 1 n}| &< \epsilon
\\
|(x^{\frac 1 n}-a^{\frac 1 n})^n| &< \epsilon ^n
\\
\left|\sum_{i=0}^{n}C(n,i)x^{\frac in}(-a)^{1-\frac i n}\right| &< \epsilon ^n
\end{align}
(it looks like a one but the power on $x$ and $a$ it's actually $\frac i n$)
if we take $\delta \leq 1$
$$
|x| < |a| + 1
\\
\left|\sum_{i=0}^{n}C(n,i)x^{\frac in}(-a)^{1-\frac i n}\right| \leq 2^n(|a| + 1)
$$
I would love to be able to say that $\delta = \min \left\{1, \frac \epsilon{2^n(|a|+1)} \right\}$ but this doesn't seem to be the case.
I also tried using the $$x^n - y^n = (x-y)(x^{n-1} + ... + y^{n-1})$$ formula but it didn't help much either.
The way I used this formula for $n=3$ for example was:
$$|\sqrt[3]x - \sqrt[3]a| < \epsilon
\\
|x-a| < \epsilon ( \sqrt[3]{x^2} + \sqrt[3]xa + \sqrt[3]{a^2})
$$
if we use the same trick $\delta \leq 1$
$$( \sqrt[3]{x^2} + \sqrt[3]xa + \sqrt[3]{a^2}) < 3\sqrt[3]{(|a|+1)^2}$$
but even if we set $\delta = \min \left\{1, 3\epsilon \sqrt[3]{(|a|+1)^2}\right\}$
it can't be shown that $|x-a|< \delta$ implies $|\sqrt[3]x - \sqrt[3]a| < \epsilon$
Prove continuity for cubic root using epsilon-delta

Hints: Let's have some ideas: for $n = 2$ we have:$$|\sqrt{x}-\sqrt{a}| = \frac{|x-a|}{\sqrt{x}+\sqrt{a}} \leq \frac{|x-a|}{\sqrt{a}},$$
and $\delta < \epsilon\sqrt{a}$ will do. For $n = 3$ we have: $$|\sqrt[3]{x}-\sqrt[3]{a}| = \frac{|x-a|}{\sqrt[3]{x^2} + \sqrt[3]{xa} + \sqrt[3]{a^2}} \leq \frac{|x-a|}{\sqrt[3]{a}^2},$$ if $a > 0$, so $\delta = \min\{a, \epsilon\sqrt[3]{a}^2\}$ solves it. Do it the same way for $a < 0$ ($a = 0$ is easy).
How it would be for $n = 4$? For $n$ even it is easier, we don't have to break anything into cases. 
Notice that I'm using all the way that: $$\begin{align}x-y &= (x^{1/n})^n-(y^{1/n})^n \\ &= (x^{1/n}-y^{1/n})(x^{(n-1)/n}+x^{(n-2)/n}y^{1/n}+\ldots+x^{1/n}y^{(n-2)/n}+y^{(n-1)/n}).\end{align}$$
