Let $n\in\Bbb N$. Show that $$ f(x) = {1\over x^n} $$ is continuous in its domain.
I've recently shown that $g(x) = x^n$ is continuous everywhere in $\Bbb R$. Now I want to do the same for $1/x^n$, however, I have some doubts regarding my reasoning below.
While proving the statement for $g(x)$ I've used the fact that: $$ |x^{n+1} - x_0^{n+1}| = |x-x_0||x^n + x^{n-1}x_0 + \cdots +x_0^{n}| $$
Now for some neighborhood of any $x$ (say $1$) we have that: $$ \exists M\in\Bbb R: |x^n + x^{n-1}x_0 + \cdots +x_0^{n}| \le M $$ So if we take $\delta = \min\{{\epsilon \over M}, 1\}$ we have: $$ |x-x_0| < \delta M < \epsilon $$
A similar approach, however, didn't work for $f(x) = 1/x^n$. After some trials, I've considered a bit different approach. I've recently shown that $1\over x$ is continuous in its domain. Namely: $$ \lim_{x\to x_0}{1\over x} = {1\over x_0} $$
Suppose: $$ \lim_{x\to x_0} {1\over x^n} = {1\over x_0^n} $$ Multiply both sides by $\lim_{x\to x_0} 1/x$: $$ \lim_{x\to x_0} {1\over x} \cdot \lim_{x\to x_0} {1\over x^n} = \lim_{x\to x_0} {1\over x} \cdot {1\over x_0^n} \iff \\ \lim_{x\to x_0} {1\over x^{n+1}} = {1\over x_0}\cdot {1\over x_0^n} = {1\over x_0^{n+1}} $$
I have a very weird feeling after applying induction. Bet it is not legal here, is it?
So my question is: How does one show $f(x)$ is continuous?
When $|x-y|<\delta <|x|/2$, then \begin{align*} \bigg|\frac{1}{x^n}-\frac{1}{y^n} \bigg|&= \frac{|y-x||x^{n-1}+\cdots +y^{n-1}|}{|x^ny^n|}\\& \leq \frac{\delta n( |x|+\delta)^{n-1} }{|x|^n (|x|-\delta)^n} \\&< \frac{n\delta 2\cdot 3^{n-1} }{ |x|^{n+1} }\end{align*}