$\epsilon - \delta$ proof that $\lim_{x \to a} \sqrt x = \sqrt a$
Solution 1:
If you take $\delta=\epsilon \sqrt{a}$ then we have that $|x-a|<\delta$ then $$|\sqrt{x}-\sqrt{a}|=\frac{|x-a|}{|\sqrt(x)+\sqrt(a)|}\leq \frac{|x-a|}{\sqrt{a}}<\epsilon$$Note this is for $a\neq0$ if $a=0$ then take $\delta=\epsilon^2$ so then $$|\sqrt{x}|<\sqrt{\epsilon^2}=\epsilon$$