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

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 $\delta_{\varepsilon}$.

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 $\delta = \epsilon,$ then $|x-a| < \delta \implies |x-a| < \epsilon$.

Instead, take
$\displaystyle \delta = \min\left(\frac{a}{2},\frac{\epsilon}{3a}\right)$.

Then, $|x - a| < \delta \implies \frac{a}{2} < x < \frac{3a}{2}$.

This implies that $|x + a| < \frac{5a}{2} < 3a$.

So, you have that $|x - a| < \delta$ and
$|x + a| < 3a$.

Therefore

$|x^2 - a^2| = |x - a| \times |x + a| < \delta \times 3a \leq \epsilon$.