$ {a_n}^{1/k} \rightarrow m^{1/k}$ Let $k$ be a fixed positive natural number and $\{a_n\}$ be a sequence of positive reals converging to $m$ then show that
$$(a_n)^{1/k} \hspace{.1cm} \text{converges to } \hspace{.1cm} m^{1/k}$$
as $n$ goes to $\infty$.
My attempt:
I need a hint to start it. 
 A: If one wishes to use the direct definition of a limit with $\epsilon$, then the proof would be a bit technical. 
Note that this argument below works only if $k$ is a positive natural number, which is what you have. If $k$ is given to be only a non-zero real number, then it is best to use the definition of a continuous map as outlined in the answer by @rwbogl. 
Proof: We have $a_n \to m$ as $n \to \infty$. For all $\epsilon > 0$, there exists $N$ such that $|a_n-m|<\epsilon$ if $n \ge N$. Now, we have 
\begin{align*}
|a_n^{\frac 1k}-m^{\frac 1k}|&=\frac{|a_n-m|}{\left|a_n^{\frac{k-1}k}+a_n^{\frac{k-2}k}m^{\frac 1k}+a_n^{\frac{k-3}k}m^{\frac 2k}+\cdots+a_n^{\frac 2k}m^{\frac{k-3}k}+a_n^{\frac 1k} m^{\frac{k-2}k}+m^{\frac{k-1}k} \right|} \\
&< \frac{\epsilon}{\left|a_n^{\frac{k-1}k}+a_n^{\frac{k-2}k}m^{\frac 1k}+a_n^{\frac{k-3}k}m^{\frac 2k}+\cdots+a_n^{\frac 2k}m^{\frac{k-3}k}+a_n^{\frac 1k} m^{\frac{k-2}k}+m^{\frac{k-1}k} \right|}.
\end{align*}
Thus, $a_n^{\frac 1k} \to m^{\frac 1k}$ as $n \to \infty$. $\blacksquare$
This argument works because the denominator $$\left|a_n^{\frac{k-1}k}+a_n^{\frac{k-2}k}m^{\frac 1k}+a_n^{\frac{k-3}k}m^{\frac 2k}+\cdots+a_n^{\frac 2k}m^{\frac{k-3}k}+a_n^{\frac 1k} m^{\frac{k-2}k}+m^{\frac{k-1}k} \right|$$ is non-zero and bounded. The denominator is bounded because each $a_n$ is bounded (because the sequence $\{a_n\}$ is convergent by hypothesis), $m$ is a finite number, and the denominator contains only a finite number of terms. Also, the denominator is non-zero since each $a_n$ is positive by hypothesis; this means precisely that at least the first term, $a_n^{\frac{k-1}k}$, is non-zero (all the other terms in the denominator would be zero if the limit of $\{a_n\}_{n=1}^\infty$ is zero, that is, $m=0$).
A: Take a look at Lemma 1, here. The gist is that, for any sequence $a_n$, if $f(x)$ is continuous on your domain, then $$\lim_{n\to \infty}f(a_n) = f(\lim_{n\to \infty}\ a_n)$$
A: Here is a proof without using continuity.
If $m = 0$, then it needs to be shown that $a_n^{1/k} \to 0$, which follows from
$$|a_n^{1/k} - 0| = |a_n|^{1/k} = |a_n - 0|^{1/k}.$$
If $m > 0$, let $b_n = a_n^{1/k}$ and $M = m^{1/k}$, then $M > 0$, it then follows that
$$|a_n - m| = |b_n^k - M^k| = |b_n - M|(b_n^{k - 1} + b_n^{k - 2}M + \cdots + M^{k - 1}) \geq |b_n - M|M^{k - 1}$$
Hence 
$$|b_n - M| \leq \frac{|a_n - m|}{M^{k - 1}}$$
and the result easily follows.
A: If $f$ is continuous then $x_n \to x \Rightarrow f(x_n) \to f(x)$
To prove that $f(x) = x^{1/k}$ is continuous, note that its inverse $g(y) = y^k$ is continuous so since $x_n \to m$ implies $y_n = f(x_n)$ is bounded one can consider a converging subsequence $y_{n_k} \to y$ now, since
$$g(y_{n_k}) = x_{n_k} \to m  \Rightarrow  g(y) = m \Rightarrow y = m^{1/k}$$
This proves that the only limit poiny of the bounded sequence $y_n$ is $m^{1/k}$. Therefore the sequence $y_n$ converges to $m^{1/k}$
