Using the $\epsilon-\delta$ definition show that $f(x) = \frac 1 {x^2}$ is a continuous function at any $x_0 = a, a > 0$ Using the $\epsilon-\delta$ definition show that $f(x) = \frac 1  {x^2}$
is a continuous function at any
$x_0 = a, a > 0$
I have expressed in the form:
$$lim_{x\to a}\frac1{x^2}=\frac1{a^2}$$
and thus, $$|f(x)-f(a)|=|\frac{1}{x^2}-\frac{1}{a^2}|=|\frac{a^2-x^2}{x^2a^2}|=|\frac{(x-a)(x+a)}{x^2a^2}| $$
let $\delta=1$, thus $$|\frac{(x-a)(x+a)}{x^2a^2}|< \frac{(x+a)}{x^2a^2}$$
I am stuck as to what I should let my delta be to finish off this proof

There were some (really) stupid errors in my other answer. 
We want 
$$|\frac{(x-a)(x+a)}{x^2a^2}|<\epsilon$$ We need to bound $x$ about $a$, that is of course what makes this problem harder.
If we set at least $\delta <a/2$  then
$$-\frac{a}{2}<x-a<\frac{a}{2}$$ and so $$\frac{a}{2}<x<\frac{3a}{2}$$
this gives 
$$|\frac{(x-a)(x+a)}{x^2a^2}|<|x-a|\frac{10}{a^3}$$ 
Now to achive
$$|\frac{(x-a)(x+a)}{x^2a^2}|<|x-a|\frac{10}{a^3} <\epsilon$$
we need only  $$|x-a|<\frac{a^3}{10} \epsilon$$
So we set $\delta=\min(a/2, \frac{a^3}{10} \epsilon)$
