I've been stuck on this problem for a while. I splitted a into $p/q$, so it would be $({X_n}^p)^{1/q}$, and I got the convergence of ${X_n}^p$ to be $x^p$ since it is just induction using the product rule, but I cannot figure our how to manipulate the definition of convergence (for each $\varepsilon>0$, there is a $N\ge k$ such that $|X_n-x| < \varepsilon$, if $n\ge N$) for something to a $n$-th root.
The closest literature I can find is this handout, but I don't understand how he went from step 1 to 2 and then to 3 in part 32 on slide 23. Any help?
This follows because $f(x)=x^a$ is continuous, ie, $f(x_n) \to f(x)$ if $x_n \to x$.
Edit: if you want to use the $\epsilon,\delta$ definition, observe that the binomial theorem gives
$$x+h \leq (\sqrt[n]{x}+\sqrt[n]{h})^n $$ $$ \sqrt[n]{x+h} \leq \sqrt[n]{x}+\sqrt[n]{h}$$ $$\sqrt[n]{x+h} - \sqrt[n]{x} \leq \sqrt[n]{h}, (*)$$
for $x,h>0$. By picking $\delta=\epsilon^n$ and using the above inequality, you should now be able to show $\sqrt[n]{x}$ is continuous.
Edit: If you are still confused, suppose $y,z>0$ and $|y-z|=h<\delta$. Then there are two cases.
If $z=y+h$, then (*) gives $\sqrt[n]{z} - \sqrt[n]{y} \leq \sqrt[n]{h} < \sqrt[n]{\delta}$.
If $y=z+h$, then (*) gives $\sqrt[n]{y} - \sqrt[n]{z} \leq \sqrt[n]{h} < \sqrt[n]{\delta}$.
In EITHER CASE, $|\sqrt[n]{z} - \sqrt[n]{y}| < \sqrt[n]{\delta}$.
In short, $$|y-z|<\delta \implies |\sqrt[n]{z} - \sqrt[n]{y}| < \sqrt[n]{\delta},$$ which should be enough to prove your result.