$|x-y|<\delta \implies |x^2-y^2|<\epsilon$

Question

Problem Statement:

Given $y\in\mathbb{R}$ and $\epsilon > 0$, show that for some $\delta>0$, if $x\in\mathbb{R}$ and $|x-y|<\delta$ then $|x^2-y^2|<\epsilon$.

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Attempt (Incomplete):

$|x^2-y^2| = |x+y||x-y| < (|x|+|y|)(|x-y|) < \underbrace{(|x-y|+2|y|)(|x-y|)}_{(1)} < \hspace{2mm} ?$

I've been told that I can take $\delta = \min(\sqrt{\frac{\epsilon}{2}}, \frac{\epsilon}{4|y|})$ with the idea being described below as continued from $(1)$:

$$(|x-y|+2|y|)(|x-y|) < |x-y|^2+2|y||x-y| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon $$

I'm not entirely convinced as it seems like $\delta$ takes $2$ minimums here?

Answer 1

That is almost entirely correct. There is a small problem: $y$ may well be $0$. So, I would define$$\delta=\begin{cases}\min\left\{\sqrt{\frac\varepsilon2},\frac\varepsilon{4\lvert y\rvert}\right\}&\text{ if }y\neq0\\\sqrt{\frac\varepsilon2}&\text{ if }y=0.\end{cases}$$Otherwise, it's fine.

Answer 2

Note that if $|x -y | < \delta$ it follows that $y-\delta < x < y + \delta$. Hence we get

$$|x^2 - y^2| = |x+y||x-y| \leq |x+y|\delta < |2y+\delta|\delta,$$

so if you put $\epsilon = |2y+\delta|\delta$ then you don't need to use the minimum.