# Prove $\lim_{x\to a} x^n = a^n$

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?

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 $$\delta|x^{n-1}+\ldots+a^{n-1}|<\delta M<\epsilon.$$
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}
• (+1) Maybe, just for the convenience of the OP, you should say how did you come out with this choice of $\delta$... Commented Nov 4, 2014 at 1:47