TITLE: Prove $\lim_{x\to a} x^n = a^n$
QUESTION [5 upvotes]: Prove that  $\lim_{x\to a} x^n = a^n$  for all natural numbers $n$ and all real numbers $a$. 
I need to prove this using the $\epsilon-\delta$ definition. I realize that $0<|x-a|<\delta$ and that $|x^n-a^n|<\epsilon$ for all $\epsilon>0$. I have factored $\left|x^n-a^n\right|$, made it smaller or equal to $\left|\delta(x^{n-1}+...+a^{n-1})\right|$, but I'm stuck after that. Any tips?

REPLY [6 votes]: We need to select $\delta$ so that $|\delta(x^{n-1}+\ldots+a^{n-1})|$ is less than $\epsilon>0$ for all $0<|x-a|<\delta$. The first step is to notice if $\delta$ has a fixed upper bounded, say $1$ ,$0<|x-a|<\delta\implies |x|<|a|+\delta\le |a|+1$, hence we know  by triangle inequality $$|x^{n-1}+\ldots+a^{n-1}|\le |x|^{n-1}+\ldots+|a|^{n-1},$$
which is bounded by some fixed positive constant $M$. So the second step is to add more control to $\delta$, so that  $$\delta|x^{n-1}+\ldots+a^{n-1}|<\delta M<\epsilon.$$

Now we transform above thought into a formal proof.
Given $\varepsilon>0$, choose $\delta=\min\left(\dfrac{\varepsilon}{2n(|a|+1)^{n-1}},1\right)>0$
$\forall x, 0<|x-a|<\delta$,then $|x|<|a|+1$, we have
$$\begin{align}
\left|x^n-a^n\right|&=\left|\delta(x^{n-1}+\cdots+a^{n-1})\right|\\
&\leq \delta\left(|x|^{n-1}+\cdots+|a|^{n-1}\right)\\
&<\delta n(|a|+1)^{n-1}<\varepsilon. 
\end{align}$$