Let $ \ \{a_n \} \ $ be a sequence of non-negative numbers such that $ \ \lim_{n \to \infty} a_n=a \in \mathbb{R} \ $ Let $ \ \{a_n \} \ $ be  a sequence of non-negative numbers such that $ \ \lim_{n \to \infty} a_n=a \in \mathbb{R} \ $. If $ \ k \in \mathbb{N} \ $ be fixed  , prove that $ \ \lim_{n \to \infty} a_n^{\frac{1}{k}}=a^{\frac{1}{k}} \ $
Answer:
we know that $ \ a_n \to 0 \ \Rightarrow \sqrt[k]{a_n} \to 0 \ $
Now since $ \ a_n \to a $ , we have $ \ a_n-a \to 0 \ $ 
But I can't proceed.
 A: It is not exactly a proof since we should prove that step
$$ \ \sqrt[k]{a_n-a} \to 0 \ \implies \ a_n^{\frac{1}{k}} \to a^{\frac{1}{k}} \ $$
We should proceed by the $\epsilon-\delta$ definition.
As a hint for a proof, suppose $a\neq 0$, we know that
$$ \lim_{n \to \infty} a_n=a \in \mathbb{R} $$
that is
$$|a_n-a|<\epsilon \quad \forall n>\bar n$$
then
$$|a_n^{\frac{1}{k}}-a^{\frac{1}{k}}|\implies |(a+\epsilon)^{\frac{1}{k}}-a^{\frac{1}{k}}|=a^{\frac{1}{k}}|(1+\epsilon/a)^{\frac{1}{k}}-1|\stackrel{Bernoulli}{\le}\frac{\epsilon a^{\frac{1}{k}}}{ka}=\bar \epsilon$$
therefore since we can choose $\epsilon$ arbitrary small also $\bar \epsilon$ can be choosen arbitrary small and then
$$\lim_{n \to \infty} a_n^{\frac{1}{k}}=a^{\frac{1}{k}}$$
A: The step you are missing:
For $t\in [0,1\rangle$ we have
$$(1-t)^k \le 1-t \le 1-t^k \implies 1-t \le (1-t^k)^{1/k}$$
For $x \ge y \ge 0$ set $t = \left(\frac{y}x\right)^{1/k} \in [0,1\rangle$ so
$$1-\frac{y^{1/k}}{x^{1/k}} \le \left(1-\frac{y}x\right)^{1/k} \implies x^{1/k} - y^{1/k} \le (x-y)^{1/k}$$
so $$\left|x^{1/k} - y^{1/k}\right| \le |x-y|^{1/k}, \quad\forall x,y \ge 0$$
Now we have
$$\left|a_n^{1/k} - a^{1/k}\right| \le |a_n - a|^{1/k} \xrightarrow{n\to\infty} 0$$
so $a_n^{1/k} \xrightarrow{n\to\infty} a^{1/k}$.
A: Let $a>0$(the case of $a=0$ is easy) and $y_n=\sqrt[k]{\frac{a_n}{a}}$.
Then $y_n^k \to 1$
Thus $|y_n^k-1|=|y_n-y||y_n^{k-1}+y_n^{k-2}+....+1|\geq |y_n-y|$
So you have the conclusion from this inequality because $y_n^m \geq 0,\forall m \in {1,2...k-1}$
