Proving $\frac{1}{x^2}$ is continuous. I wanted to prove that the function $f:\mathbb{R}\rightarrow\mathbb{R}$ given by $f(x)=\frac{1}{x^2}$ is continuous for all $x\in\mathbb{R}$ excluding x=0 of course. The proof goes as follows.
Let $\epsilon>0$ be given arbitrary and choose $\delta=min(1,\frac{c}{2},\frac{c^4\epsilon}{4(1+2|c|)})$. Assume that $\forall x.c\in\mathbb{R}: 0<|x-c|<\delta$. It follows:
$$|f(x)-f(c)|=|\frac{1}{x^2}-\frac{1}{c^2}|=|\frac{x^2-c^2}{x^2c^2}=|\frac{|x-c||x+c|}{x^2c^2}|=|\frac{|x-c||(x-c)+2c|}{x^2c^2}|\leq\frac{|x-c|(|x-c|+2|c|)}{x^2c^2}$$ $$ \leq\frac{|x-c|(1+2|c|)}{x^2c^2}<\frac{|x-c|4(1+2|c|)}{c^2c^2}<4\delta\frac{1+2|c|}{c^4}\leq\epsilon$$
Is this proof valid and in particular is the delta I chose fine and if not provide some tips please.
 A: Let $\epsilon>0$ is arbitrary. We will consider $c>0$ (for $c<0$ we can follow similar logic).
$$|f(x)-f(c)|=|\frac{1}{x^2}-\frac{1}{c^2}|=|\frac{x^2-c^2}{x^2c^2}|=|x-c|\frac{|x+c|}{x^2c^2}$$ From here, we need to find bounds for $x \text{ and }|x+c|$ than do not depend on $x$. Let $\delta_1=0.5c$. Then $|x-c|<\delta_1 \implies 0.5c < x < 1.5c, 1.5c <x+c < 2.5c$ We have for $|x-c|<\delta_1$:
$$|f(x)-f(c)|=|x-c|\frac{|x+c|}{x^2c^2} < \delta_1 \frac{2.5c}{0.25c^4}=\frac{10\delta_1}{c^3}$$
Next, pick $\large{\delta=min\{\delta_1, \frac{\large{c^3}\epsilon}{10}\}}$  then for $|x-c| < \delta$ we have $|f(x)-f(c)| < \epsilon$
A: To give a different perspective:
If you know that the composition of two continuous functions is continuous, then the problem reduces to showing that $x\mapsto x^2$ is continuous on $\mathbb{R}$ and $x\mapsto \frac{1}{x}$ is continuous on $\mathbb{R}\setminus\{0\}$. This is not necessarily easier than showing it directly, but you might have shown these two things beforehand.
