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.

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.
