Show that $x^n$ is a continuous function So the idea I thought of doing was by induction.
$\forall\; \epsilon > 0\;\; \exists \;\delta > 0$ such that $|x-x_0| < \delta \implies |x-x_0| < \epsilon $
By the definition of the limit choose $\delta=\epsilon$ and the statement holds for $n=1$. Thus $\lim_{x \rightarrow x_0} = x_0 = f(x_0)$
Now here is the inductive step.
Next assume that $\lim_{x \rightarrow x_0} x^n$ exists. Another words, $\forall\; \epsilon > 0\;\; \exists \;\delta > 0$ such that $|x-x_0| < \delta \implies |x^n-x_0^n| < \epsilon $
$\forall\; \epsilon > 0\;\; \exists \;\delta > 0$ such that $|x-x_0| < \delta \implies |x^{n+1}-x_0^{n+1}| < \epsilon $
Let $\epsilon >0$ then $|x-x_0| < \delta \implies |x-x_0|^n|x+x_0| < \epsilon$ 
I am having trouble proceding from here any hints or advice would be greatly appreciated.

Hint: if $f$ is continuous and $g$ is continuous then $fg$ is continuous. This is not that hard to show (ask if you need help). Then apply induction.
