$\lim_{x \to a} x^2 = a^2$. As per the definition of limits if $\lim_{x \to a} f(x)= L$, then $$\forall \varepsilon \gt 0 \ \exists \delta \gt 0 \ s.t 0\lt\lvert x-a \rvert \lt \delta \ \implies \ 0\lt \lvert f(x)- L\rvert \lt \varepsilon $$
I want to prove that $\lim_{x \to a} x^2 = a^2$.
As per the definition $$\lvert f(x)- L\rvert = \lvert x^2- a^2\rvert = \lvert (x-a)(x+a)\rvert =\lvert x-a\rvert \lvert x+a \rvert$$
As per definition $$\lvert x-a \rvert \lt \delta \implies -\delta \lt x-a \lt \delta \implies a-\delta \lt x <a+\delta \implies 2a-\delta \lt x+a <2a+\delta $$
I'm stuck beyond this. I cannot find a suitable $\varepsilon$ to satisfy my condition here.

You must find a solution to the equation in $\delta$
$$|x-a|<\delta\implies|x^2-a^2|<\epsilon.$$
The RHS is
$$\sqrt{a^2-\epsilon}<x<\sqrt{a^2+\epsilon}$$ or
$$\sqrt{a^2-\epsilon}-a<x-a<\sqrt{a^2+\epsilon}-a.$$
So if we take
$$0<\delta<\min(a-\sqrt{a^2-\epsilon},\sqrt{a^2+\epsilon}-a)$$ the conditions are fulfilled.
Note that if $\epsilon\ge a^2$, the condition $a^2-x^2<\epsilon$ is automatically fulfilled by $0\le x\le a$.
