# 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.

• Your title and body disagree. (The $1/n$ in the exponent in the title is probably the error.) – Thomas Andrews Oct 5 '17 at 19:02
• Do you know that $x^{1/k}=e^{\log(x)/k}$, and that the exponential function and the natural logarithm are continuous? – Henning Makholm Oct 5 '17 at 20:14

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.
• This is perhaps the simplest approach. +1. We also need to cover cases when $\lim x_{n} ^{1/k}$ does not exist and show that in these cases $\lim x_{n}$ wouldn't exist. – Paramanand Singh Oct 6 '17 at 1:52