TITLE: $\epsilon - \delta$ proof that $\lim_{x \to a} \sqrt x = \sqrt a$
QUESTION [6 upvotes]: I am trying to prove, 
$\lim_{x \to a} \sqrt x = \sqrt a$
As per the definition of limit for every $\epsilon > 0$ there is some $\delta > 0$ such that 
 $ 0 < |x-a| < \delta$ implies $|f(x) - \sqrt a| < \epsilon$
$|\sqrt x - \sqrt a|  = \frac {|x -a|}{\sqrt x + \sqrt a} $
since 
$ 0 < |x-a| < \delta$
$\frac {|x -a|}{\sqrt x + \sqrt a} < \delta$
Can I stop my proof at this point since I found $\epsilon$ ( which is $\delta$ in this case)

REPLY [4 votes]: 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 $$|\sqrt{x}|<\sqrt{\epsilon^2}=\epsilon$$