Question 15 of Chapter 1 of Pugh's Real Mathematical Analysis, Ed. 2
Given $y \in \mathbb{R}$, $n \in \mathbb{N}$, and $\epsilon > 0$ show that for some $\delta > 0$, if $u \in \mathbb{R}$ and $|u-y| < \delta$ then $|u^n-y^n| < \epsilon.$
Hint: Prove the inequality when $n=1$, $n=2$, and then do induction on $n$ using the identity $u^n - y^n = (u-y)(u^{n-1} + u^{n-2}y + \dots +y^{n-1}).$
The question seems to be asking that the function $u \mapsto |u^n -y^n|$ is continuous for all $y,n$. I don't quite understand why the hint is to use induction. Can this not be solved by direct means?
Namely, if $y < u < (y^n + \epsilon)^{1/n}$, then $u^n - y^n < (y^n + \epsilon) - y^n = \epsilon$, which gives the required $\delta$. Is there something wrong the proof that I have outlined, so that induction is needed?
The case $n=2$ follows from
$$|u^2 - y^2|=|u-y||u+y|\leq |u-y|(|u| + |y|) \leq |u-y|(2|y| + |u - y|).$$
Similarly, we see that
$$|u^n - y^n|=|u-y|\left|\sum_{k=1}^{n}{y^{n-k} u^{k-1}}\right|$$
and
$$\left|\sum_{k=1}^{n}{y^{n-k}u^{k-1}}\right|\leq\sum_{k=1}^{n}{|y^{n-k}||u^{k-1}|}\leq\sum_{k=1}^{n}{|y^{n-k}|(|y^{k-1}|+|u^{k-1}-y^{k-1}|)}.$$
Now the conclusion follows from induction.