I'm attempting to prove the following: Fix $\alpha \in (0,1)$, and let $\{x_{n}\}_{n \ge 0}$ be a sequence in $[0,\infty)$ such that $\lim_{n \rightarrow \infty}x_{n} = L$. Prove $\lim_{n \rightarrow \infty}x_{n}^{\alpha} = L^{\alpha}$.
Obviously if I can show $|x_{n}^{\alpha}-L^{\alpha}| \le |(x_{n}-L)|^{\alpha}$ then I am done. How can I go about proving this smaller statement? A hint would be preferred. Or do I need to take another approach all together?
Notice that $L<(L^{\alpha}+\epsilon)^{\frac{1}{\alpha}}$ and $L>(L^{\alpha}-\epsilon)^{\frac{1}{\alpha}}$. Now given that $\lim_{n\rightarrow \infty }x_n=L$, there exists some $N$ such that $x_n<(L^{\alpha}+\epsilon )^{\frac{1}{\alpha }}$ and $x_n>(L^{\alpha}-\epsilon )^{\frac{1}{\alpha }}$ for $n \geq N$.
Then we get by taking power $\alpha$:
$x_n^{\alpha }>(L^{\alpha}-\epsilon )$ and $x_n^{\alpha }<(L^{\alpha}+\epsilon )$.