Proving the function $f(x)=x^n$ is continuous

Question

I want to prove that given $y\in \mathbb R, n\in \mathbb N, \epsilon >0$, there exists $\delta>0$ such that for any $u\in\mathbb R$, $|u-y|<\delta$ implies $|u^n-y^n|<\epsilon$. It is not permitted to use any notion of product rule of limits, continuous function etc. Only a pure epsilon-delta argument has to be used.

So here's my attempt:

We proceed by induction. For $n=1$, clearly $\delta=\epsilon$. Assume the result true for all $k\le n-1$. Now fix $\epsilon>0$ and consider $|u^n-y^n|=|u-y||u^{n-1}+u^{n-2}y+\cdots +y^{n-1}|$. The term $|u^{n-1}+u^{n-2}y+\cdots +y^{n-1}|$ seems to be the issue. I need to bound this quantity in terms of $\epsilon$ by using the induction hypothesis. Now I am thinking of something like $|u^{n-j}|<\frac{\epsilon}{n}+|y^{n-j}|$ but cannot seem to get a coherent argument.

Answer 1

If $|y-x|<1$, then $|y|=|y-x+x|\leqslant|x|+1$. Therefore\begin{align}|x^{n-1}+x^{n-2}y+\cdots+y^{n-1}|&\leqslant|x|^{n-1}+|x|^{n-2}\bigl(|x|+1\bigr)+\cdots+\bigl(|x|+1\bigr)^{n-1}\\&\leqslant n\bigl(|x|+1\bigr)^{n-1}.\end{align}So, take$$\delta=\min\left\{1,\frac\varepsilon{n\bigl(|x|+1\bigr)^{n-1}}\right\}.$$

Answer 2

HINT Let's take a specific example, say $n=2$. You need to argue, given $y \in \mathbb{R}$ and $\epsilon > 0$ that there exists a $\delta > 0$ such that $\left|y^2-u^2\right| < \epsilon$ for all $u \in (y-\delta,y+\delta)$.

So if $u \in (y-\delta,y+\delta)$, then we have $$ \left|y^2-u^2\right| = |u-y||u+y| \le \delta (2y+\delta) $$ and so you need to pick $\delta$ so that $\delta (2y+\delta) < \epsilon$.