$\lim_{x \to a} x^2 = a^2$. 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: 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 $\delta_{\varepsilon}$.
A: 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 $0\le x\le a$.
A: 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
$|x^2 - a^2| = |x - a| \times |x + a| < \delta \times 3a \leq \epsilon$.
A: For all real numbers $x$ and $a$ we have
\begin{eqnarray}
|x^2-a^2|&=&|(x-a)(x+a)|\cr
&=&|x-a||x+a|\cr
&=&|x-a||(x-a)+a|\cr
&\le&|x-a|(|x-a|+|a|)\cr
&=& |x-a|^2+|a||x-a|
\end{eqnarray}
If $\varepsilon>0$ is such that
$$
|x-a|^2+|a||x-a|<\varepsilon,
$$
then
$$
\frac{-|a|-\sqrt{a^2+4\varepsilon}}{2}<|x-a|<\frac{-|a|+\sqrt{a^2+4\varepsilon}}{2}
$$
Hence, if we choose
$$
\delta=\frac{-|a|+\sqrt{a^2+4\varepsilon}}{2}=\frac{2\varepsilon}{|a|+\sqrt{a^2+4\varepsilon}}>0
$$
for all $x$ satisfying $|x-a|<\delta$, we have $|x^2-a^2|<\varepsilon$
