Proof that $\lim (x_n)^{1/k}=(\lim x_n)^{1/k}$ I'm trying to prove $\lim (x_n)^{1/k}=(\lim x_n)^{1/k}$ having already proved it for $k=2$.  Even experimenting with $k=3$ I'm getting stuck.  If we call $x=\lim x_n$ then I have 
$$|(x_n)^{1/3}-x^{1/3}| \cdot \left|\frac{(x_n)^{2/3}+x^{2/3}}{(x_n)^{2/3}+x^{2/3}}\right| = \left|\frac{x_n-x-x_n^{2/3}x^{1/3}+x_n^{1/3}x^{2/3}}{(x_n)^{2/3}+x^{2/3}}\right|$$
In the proof for $k=2$ I didn't have this positive term in the numerator to deal with.  I know that I could choose $|x_n-x|<\min\{\varepsilon,x/2\}$ which would imply that $x<2x_n$ but this seems to suggest a way to replace $x$ rather than $x_n$ which seems unhelpful.

Do not complicate things, just use the fact that if $a_n$ and $b_n$ are convergent then $a_nb_n$  also is and $\lim a_n b_n=\lim a_n \cdot \lim b_n$.
Suppose that $\lim ((x_n)^{1/k}) \neq (\lim x_n)^{1/k}$ and take both sides to the power of $k$ and use above written law and then conclude. 
