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
 A: Saying $\lim_{x → a} 1/x² = 1 / a²$ , for $a > 0$, is, by definition the same as
$$
∀ ε > 0 . ∃ δ > 0 . ∀ x . |x - a| < δ ⇒ |1/x² - 1/a²| < ε
$$
So suppose we're given $a > 0$ and $ε > 0$,
then we are tasked with finding a $δ > 0$ that satisfies the above property. I'm no clever guy, so the best I can do is start with the property that it needs to satisfy and manipulate that with the aim of a candidate $δ$ popping-up. Let's try that:
  | 1 / x² - 1 / a² | < ε
≡⟨ multiplication ⟩
  | a² / a²x² - x² / x²a² | < ε
≡⟨ fractions ⟩
  | (a² - x²) / (ax)² | < ε
≡⟨ difference of squares ⟩
  |(a + x)(a - x)/ (ax)²| < ε
≡⟨ absolute value ⟩
  |a + x||a - x|/ (ax)² < ε
≡⟨ algebra ⟩
  |a - x| < ε · (ax)² · |a + x|

So it seems if we take
$$ δ = ε · (ax)² · |a + x|$$
which is positive since each factor is positive, then we can show
---rather have already shown--- that
|1/x² - 1/a²| < ε.
Hope that helps :-)
A: 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 $\delta=\min(a/2, \frac{a^3}{10} \epsilon)$
