$\lim_{x\to a}{x^2=a^2}$
I get stuck midway, maybe i should give values to $\delta$?
Here is the process
$\left|x^2-a^2\right|<\epsilon \Rightarrow \left|x-a\right|\left|x+a\right|<\epsilon$ then i know that $\left|x-a\right|<\delta$ so $\delta\left|x+a\right|$<$\epsilon $ now i do not know what's next
Hint 1: $x^2-a^2=(x-a)(x+a)$. You have control over $|x-a|$ directly with $\delta$. Try to bound $|x+a|$.
Hint 2: Taking $\delta<1$ can make things cleaner.
Solve the inequality $|x^2-a^2|<\varepsilon$:
$$ \begin{cases} x^2-a^2&<\varepsilon\\ x^2-a^2&>-\varepsilon \end{cases} $$
By solving both inequalities, you arrive at
$$ \begin{cases} -\sqrt{a^2+\varepsilon}<x<\sqrt{a^2+\varepsilon}\\ x<-\sqrt{a^2-\varepsilon}\;\vee\;x>\sqrt{a^2-\varepsilon} \end{cases} $$
(without loss of generality for the second inequality).
Then the system is solved by
$$ -\sqrt{a^2+\varepsilon}<x<-\sqrt{a^2-\varepsilon} \;\vee\; \sqrt{a^2-\varepsilon}<x<\sqrt{a^2+\varepsilon}. $$
From the second part you can conlude that there exists a number $\delta$ such that, for all $x$, where $0<|x-a|<\delta$, you have that $|x^2-a^2|<\varepsilon$: just take $\delta=\min\{|\sqrt{a^2+\varepsilon}-a|,|a-\sqrt{a^2-\varepsilon}|\}$.
