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.
Best Answer
Induction isn't really necessary here. It is enough to just work with arbitrary $n$ to show the argument works for whatever $n$ you choose.
Indeed, fix $\varepsilon > 0$. We want $|x^{n+1} - x_0^{n+1}| < \varepsilon$. Notice that
$$|x^{n+1} - x_0^{n+1}| = |x - x_0||x^n + x^{n-1}x_0 + \dots + x x_0^{n-1} + x_0^n|$$
When $x$ is in an interval around $x_0$, say $x \in [x_0 - 1, x_0+1]$, the sum on the RHS has a bound, call this bound $M$.
Hence, taking $\delta = \min(\dfrac{\varepsilon}{M}, 1)$ we get
$$|x^{n+1} - x_0^{n+1}| = |x - x_0||x^n + x^{n-1}x_0 + \dots + x x_0^{n-1} + x_0^n| < \delta M = \varepsilon$$
so we are done.