Joshua Foley

2022-07-01

As per the definition of limits if $\underset{x\to a}{lim}f\left(x\right)=L$, then

poquetahr

A different approach for the sake of curiosity.
Let $0<|x-a|<{\delta }_{\epsilon }$. Then we have that:
$\begin{array}{rl}|f\left(x\right)-L|& =|{x}^{2}-{a}^{2}|\\ \\ & =|\left(x-a\right)\left(x+a\right)|\\ \\ & =|x-a||\left(x-a\right)+2a|\\ \\ & \le |x-a|\left(|x-a|+2|a|\right)\\ \\ & <{\delta }_{\epsilon }\left({\delta }_{\epsilon }+2|a|\right):=\epsilon \end{array}$
where you can choose the positive root of the corresponding equation on ${\delta }_{\epsilon }$

uplakanimkk

Alternative approach:
Without loss of generality, $a>0$. That is, the approach for $a<0$ is similar, while the approach if $a=0$ is trivial.
If $\delta =ϵ,$ then $|x-a|<\delta \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|x-a|<ϵ$
Instead, take $\delta =min\left(\frac{a}{2},\frac{ϵ}{3a}\right)$
Then, $|x-a|<\delta \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{a}{2}
This implies that $|x+a|<\frac{5a}{2}<3a$
So, you have that $|x-a|<\delta$ and$|x+a|<3a$
Therefore
$|{x}^{2}-{a}^{2}|=|x-a|×|x+a|<\delta ×3a\le ϵ$

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