# $\epsilon - \delta$ proof that $\lim_{x \to a} \sqrt x = \sqrt a$

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)

• Can you rewrite what you're trying to prove, the $\epsilon$-$\delta$ part? As written, it's not quite right. – davidlowryduda Mar 7 '14 at 23:08
• You have partly interchanged $\epsilon$ and $\delta$. Considerable modification is needed for a proof. The one item that will be useful is the expression $\frac{|x-a|}{\sqrt{x}+\sqrt{a}}$. – André Nicolas Mar 7 '14 at 23:09
• @mixedmath oops. I corrected it. – Surya Mar 7 '14 at 23:20
• @AndréNicolas I am not sure what you mean. – Surya Mar 7 '14 at 23:21
• The interchanging happened earlier, your modified post does not interchange. – André Nicolas Mar 7 '14 at 23:40

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 $$|\sqrt{x}|<\sqrt{\epsilon^2}=\epsilon$$
• Hi Kaladin, thank you. I want to know why I can't stop at the point that I stopped and call my proof done. I already proved that $|\sqrt x - \sqrt a| < \delta$, I found my $\epsilon$ why can't i stop there. what I am I missing here. – Surya Mar 7 '14 at 23:42
• You said since $0<|x-a|<\delta$ we can say $\frac{|x-a|}{\sqrt{a}+\sqrt{x}}<\delta$ which doesn't has to be true if $\sqrt{a}+\sqrt{x}<1$ – Kaladin Mar 7 '14 at 23:52
• example $x=1/4$ so $\sqrt{x}=1/2$ and $a=1/16$ so $\sqrt{a}=1/4$ then $|x-a|=3/16<4/17$ but $|\sqrt{x}+\sqrt{a}|=3/4$ and $(3/16)/(3/4)=1/4>4/17$ – Kaladin Mar 7 '14 at 23:55
• The value of delta you choosed dose not guarantee that the x's in this interval are positive, so it's possible that $\sqrt(x)$ is not even defined in this interval. take epsilon =1, and a<1. You need to put another constriction that depends on a. Like : let delta less than $\frac{a}{2}$, that should solve the problem. – yousef magableh May 9 at 18:59