Prove $$\lim_{n\to\infty} x_n^{1/k} = \left(\lim_{n\to\infty} x_n\right)^{1/k}.$$
$\lim x_n = x$, and $x_n \ge 0$, $\varepsilon >0$
when $x=0$, $|x_n-x| \lt \varepsilon^k$ $\rightarrow$ $|x_n|< \varepsilon^k$ $\rightarrow$ $x_n^{\frac 1k} \lt \varepsilon$.
when $x \gt 0$, $|x_n^{\frac 1k}-x^{\frac 1k}|= \dfrac {|x_n-x|}{x_n^\frac{k-1}{k}+x^\frac{k-1}{k}}\le \dfrac {|x_n-x|}{x^\frac{k-1}{k}} $. Suppose $|x_n-x| \lt \varepsilon\cdot x^\frac{k-1}{k}$. Then, $|x_n^{\frac 1k}-x^{\frac 1k}|<\varepsilon$.
Could you tell me whether the proof is valid??
Thank you in advance.
Your $x=0$ is fine as long as you make some mention of "there exists some $N$ so that for all $n\geq N$, $|x_n-x|<\epsilon^k$.
To deal with $x > 0$ case, let $y_n = x_n^{1/k}$, $x=y^{1/k}$, then
$$y_n^k - y^k = (y_n-y)(y_n^{k-1}+y_n^{k-2}y + \dotsb + y^{k-1})$$ Rearranging and rewriting in terms of $x_n$, and noting that all terms are non-negative
$$|x_n^{1/k}-x^{1/k}| = \frac{|x_n-x |}{x_n^{1-1/k} + x_n^{1-2/k}x^{1/k} + \dotsb + x_n^{1/k}x^{1-2/k} + x^{1-1/k}}$$ Now, fix any $m$ that satisfies $0<m<x$. Given that $x_n\to x$, it follows that there must exist some $N_1$ so that $m < x_n$ for all $n\geq N_1$ (as $x_n$ eventually gets arbitrarily close to $x$).
Pick $\epsilon > 0$, as $x_n \to x$ there exists some $N_2$ so that for $n\geq N_2$, $$|x_n-x| \leq \epsilon m^{1-1/k} k$$ hence if $n \geq \max\{N_1,N_2\}$ then
$$|x_n^{1/k}-x^{1/k}| < \frac{|x_n-x|}{m^{1-1/k}+m^{1-2/k}m^{1/k}+\dotsb + m^{1-1/k}} = \frac{|x_n-x|}{km^{1-1/k}} < \epsilon $$