Proving that $\frac{1}{x^2}$ is continuous on (0,1)
$\forall \epsilon >0$ , $ \exists \delta>0 $ such that $\lvert x-x_0 \rvert < \delta $ and $\lvert f(x)-f(x_0) \rvert < \epsilon$
So can I take $\lvert\frac{1}{x^2}-\frac{1}{x_0^2} \rvert < \epsilon$
$\lvert\frac{ x_0^2 -x^2}{x_0^2 x^2 } \rvert=\lvert \frac{(x_0-x)(x_0+x)}{x_0^2 x^2} \rvert = \lvert x-x_0 \rvert \lvert \frac{x+x_0}{x_0^2 x^2}\rvert < \delta* \frac{x+x_0}{x_0^2 x^2}< \epsilon$
I'm confused as to what to do now. Am I looking to create my $\delta$ function or do I have to examine individual cases?