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.
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