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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?

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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 $$\begin{align} \left|x^n-a^n\right|&=\left|\delta(x^{n-1}+\cdots+a^{n-1})\right|\\ &\leq \delta\left(|x|^{n-1}+\cdots+|a|^{n-1}\right)\\ &<\delta n(|a|+1)^{n-1}<\varepsilon. \end{align}$$

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    $\begingroup$ (+1) Maybe, just for the convenience of the OP, you should say how did you come out with this choice of $\delta$... $\endgroup$
    – ILoveMath
    Commented Nov 4, 2014 at 1:47

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