If $\lim_{n \to \infty} a_n=L$, then $\lim_{n \to \infty} a_n^{1/k}=L^{1/k}$. {Edited] I need some help proving the following statement:
[EDIT]: forgot to mention that:
$a_n \ge 0$ , $L \ge 0$.
If the limit of a sequence $\lim_{ \ n \to \infty} a_n=L$,
then, for any $k \in \mathbb{N}$, the limit of the sequence $\lim_{n \to \infty} \sqrt[k]{a_n}  = \sqrt[k]{L}$. 
I already proved that every sequence $\{a_n\}^k$ converges to $L^k$ by induction step,
but when I tried the same for $\{1/k\}$, it just doesn't work.
 A: Let $a_{n} \geq 0$ for all $n \geq 1$, so that, if $a_{n} \to L$, then $L \geq 0$. This allows $a_{n}^{1/k}$ to be meaningful for all $n,k \geq 1$.
Let $k \geq 1$. If $L=0$, then the statement is obvious; for, we have $a_{n} < \varepsilon$ iff $a_{n}^{1/k} < \varepsilon^{1/k}$. Suppose $L > 0$. Then there is some $N_{1} \geq 1$ such that $a_{n} > 0$ for all $n \geq N_{1}$. If $n \geq N_{1}$, then
$$
|a_{n}^{1/k} - L^{1/k}| = \frac{|a_{n}-L|}{ \sum_{j=0}^{k-1}a_{n}^{\frac{j}{k}}L^{\frac{k-1-j}{k}}}.
$$
On the other hand, there is some $N_{2} \geq N_{1}$ such that $n \geq N_{2}$ implies
$|a_{n} - L| < L/2$, implying $L/2 < a_{n}$, implying that
$$
\sum_{j=0}^{k-1} a_{n}^{\frac{j}{k}} L^{\frac{k-1-j}{k}} >  L^{\frac{k-1}{k}}/ \sum_{j=0}^{k-1}{2^{\frac{j}{k}}} =: M.
$$
If $\varepsilon > 0$, then there is some $N_{3} \geq N_{2}$ such that $n \geq N_{3}$ implies $|a_{n}-L| < M\varepsilon$.
We have proved that, for every $\varepsilon > 0$, there is some $N \geq 1$ such that $n \geq N$ implies $|a_{n}^{1/k} - L^{1/k}| < \varepsilon$.
A: You can for example use the fact that $f(x) = x^p$ is continuous together with the definition of limit (and continuity), it's straight forward just to chain these definitions.
The only thing that may remain is to prove that $f$ is continuous. This is done by providing an estimate of $f(x+h)-f(x)$, that is to become arbitrarily small for small enough $h$. We can use monotonicity of $f$ and that we can compute it's inverse. Just assume $\delta>0$ and consider $f^{-1}(f(x)\pm\delta) = (x^p\pm\delta)^{1/p}$, this gives us an interval such that $|f(x+h)-f(x)|<\delta$.
To be concrete assume that $k$ is a positive integer, and that $\lim a_n=L$. Consider a $\delta > 0$, now consider any $x$ such that $(L^{1/k}-\delta)^k <x<(L^{1/k}+\delta)^k$, then we will have $L^{1/k}-\delta<x^{1/k}<L^{1/k}+\delta$. Now since $\lim a_n = L$ and that $(L^{1/k} - \delta)^k < (L^{1/k})^k = L < (L^{1/k}+\delta)^k$ we have a $N$ such that $|a_n - L|$ is less than both $L - (L^{1/k}-\delta)^k$ and $(L^{1/k}+\delta)^k-L$ and there fore $(L^{1/k}-\delta)^k < a_n < (L^{1/k}+\delta)^k$ for all $n>N$, therefore we have $L^{1/k}-\delta<a_n^{1/k}<L^{1/k}+\delta$ (which shows that $\lim a_n^{1/k} = L^{1/k}$). Note though that we must take special care if $L^{1/k}-\delta<0$, for $L>0$ this means only to restrict $\delta$ so this can't happen, but in general we would use the fact that $a_n>0$ instead for the lower bound in the inequalities.
