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.
 A: 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.
A: 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.
A: Your proof by induction would be less clunky if you indexed each $\epsilon$ in your proof. This would simplify matters. Since you are doing this proof concerning $f_n:\mathbb{R}\rightarrow\mathbb{R},f_n(x)=x^n$ for every $n\in\mathbb{Z}^+$, you may want to use $\epsilon_n$ for each $f_n$. So for $n=1$, you should have $$\forall\epsilon_1\gt0,\exists\delta=\epsilon_1,\forall{x},|x-x_0|\lt\delta\implies|x-x_0|\lt\epsilon_1.$$ For $n=k$, you should have $$\forall\epsilon_k\gt0,\exists\delta,\forall{x},|x-x_0|\lt\delta\implies|x^k-x_0^k|\lt\epsilon_k.$$ However, I should also point out that induction is actually very unhelpful for this proof, and it will lead you effectively nowhere. Instead, consider this, $$\forall\epsilon\gt0,\exists\delta,\forall{x},|x-x_0|\lt\delta\implies|x^n-x_0^n|\lt\epsilon$$ and notice that $$\left|x^n-x_0^n\right|=|x-x_0|\left|\sum_{m=0}^{n-1}x^mx_0^{n-m-1}\right|\leq|x-x_0|\sum_{m=0}^{n-1}|x|^m|x_0|^{n-m-1}=|x-x_0|\sum_{m=0}^{n-1}|(x-x_0)+x_0|^m|x_0|^{n-m-1}\leq|x-x_0|\sum_{m=0}^{n-1}(|x-x_0|+|x_0|)^m|x_0|^{n-m-1}\lt\delta\sum_{m=0}^{n-1}(\delta+|x_0|)^m|x_0|^{n-m-1}.$$ As such, let $\epsilon=\delta\sum_{m=0}^{n-1}(\delta+|x_0|)^m|x_0|^{n-m-1}$. What remains is to prove that whenever $\delta\sum_{m=0}^{n-1}(\delta+|x_0|)^m|x_0|^{n-m-1}\gt0$, there exists a corresponding $\delta\gt0$. This is just algebra, and so it should not be difficult for you to prove.
