Using delta epsilon to prove the continuity of a function 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? 
 A: To give an upper bound to
$$\frac{x+x_0}{x_0^2 x^2}$$
first we have $x_0, x\in (0,1)$ so 
$$\frac{x+x_0}{x_0^2 x^2}\le \frac{2}{x_0^2 x^2}. $$
But we still want to give a bound to $x$ (we do not need to care about $x_0$ as it is fixed). So we need a lower bound for $x$ (thus an upper bound for $1/x^2$). Then we make an extra assumption that $x > x_0/2$ (or $|x-x_0| <x_0/2$). Then 
$$\frac{x+x_0}{x_0^2 x^2} \le \frac{8}{x_0^4}\Rightarrow \left| \frac{1}{x^2} - \frac{1}{x_0^2}\right| \le \frac{8}{x_0^4} |x-x_0|$$
Now it is clear how we choose $\delta$. For all $\epsilon >0$, choose 
$$\delta = \min\left\{ \frac{x_0^4}{8} \epsilon, \frac{x_0}{2}\right\}.$$
Then whenever $|x-x_0|<\delta$, we have
$$\left| \frac{1}{x^2} - \frac{1}{x_0^2}\right| \le \frac{8}{x_0^4} |x-x_0| < \epsilon.$$
(In the first inequality we used $\delta <x_0/2$, in the second we used $\delta < \frac{x_0^4}{8} \epsilon$). As $\epsilon >0$ is arbitrary, we have 
$$\lim_{x\to x_0} \frac{1}{x^2} = \frac{1}{x_0^2}$$
and so $\frac{1}{x^2}$ is continuous on $(0,1)$. 
