$\lim_{x \to a} x^2 = a^2$. 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.

A different approach for the sake of curiosity.
Let $0 < |x - a| < \delta_{\varepsilon}$. Then we have that:
\begin{align*}
|f(x) - L| & = |x^{2} - a^{2}|\\\\
& = |(x-a)(x+a)|\\\\
& = |x - a||(x - a) + 2a|\\\\
& \leq |x - a|(|x - a| + 2|a|)\\\\
& < \delta_{\varepsilon}(\delta_{\varepsilon} + 2|a|) := \varepsilon
\end{align*}
where you can choose the positive root of the corresponding equation on $\delta_{\varepsilon}$.
