Find the value of $\delta$?

Question

Let $a>0$ and $\epsilon >0$. Determine $\delta$ (depend on $a$ and $\epsilon$) such that for all $x>0$, $\vert x-a \vert < \delta$ implies $\vert 1/x^2 - 1/a^2 \vert <\epsilon$.

I tried and found $\delta = 3 a^3 \epsilon$. Is my ans correct? Please help

Answer 1

Let $x >0$, $a>0$, and $\epsilon >0$ be given.

Consider $| x-a| < a/2$. Then $a/2 < x < (3/2)a$

Choose $\delta \le \min(a/2,(\dfrac{a^3}{10})\epsilon)$.

Then $|x-a| < \delta$ implies

$|1/x^2-1/a^2|= |\dfrac{(x-a)(x+a)}{x^2a^2}| $

$\lt \dfrac{|x-a|(5/2)a}{(a^4/4)} \lt \delta (\dfrac{10}{a^3}) \lt \epsilon.$

Answer 2

Note that such a $\delta$ is not uniquely determined. You have done some hidden calculations and finally came up with the proposal $$\delta=3a^3\epsilon\ .$$ This answer is correct iff $|x-a|<\delta$ implies $\bigl|1/ x^2-1/a^2\bigr|<\epsilon$ under all circumstances. If it is correct a proof will not be easy. Therefore I try to give a counterexample.

Assume that the givens are such that $3a^2\epsilon<1$, and put for simplicity $$x:=a-3a^3\epsilon>0\ .$$ Then $$\left|{1\over x^2}-{1\over a^2}\right|={|x-a|(a+x)\over a^2 x^2}\geq{3a^3\epsilon(2-3a^2\epsilon)a\over a^4}\geq3\epsilon\ .$$ This busts your proposal.