Prove $\lim_{x\to a}\sqrt{x}=\sqrt{a}$ by definition Stewart solution error Let $a>0$. I want to show
$$\lim_{x\to a}\sqrt{x}=\sqrt{a}$$
Purpose of this question was to investigate Stewart's online solution, which turns out to be incorrect.
Their solution
I don't understand the solution on their website. Are they saying that
$1/x  < x $? Seems that way to me by the way they choose delta.
My solution :
$$|x-a|< a   \implies   \frac{1}{\sqrt{x} + \sqrt{a}} < \frac{1}{\sqrt{a}}$$
and choose delta accordingly. Is that correct?
Thanks for the help.
 A: You have to discuss the inequality $|\sqrt{x}-\sqrt{a}|<\varepsilon$. 
Now take $\delta=\min(\varepsilon\sqrt{a},a)$ and assume $|x-a|<\delta$. Since $\sqrt{x}+\sqrt{a}>\sqrt{a}$, we have
$$
|\sqrt{x}-\sqrt{a}|=
\frac{|x-a|}{\sqrt{x}+\sqrt{a}}<
\frac{|x-a|}{\sqrt{a}}<\frac{\delta}{\sqrt{a}}\le\varepsilon
$$
A: Here is the outline for this style of proof:
Prove 
$\lim_\limits {x\to a} \sqrt x = \sqrt a$
What is the definition of "limit"?
$\forall \epsilon>0, \exists \delta >0$ such that $|x-a|< \delta \implies |\sqrt x - \sqrt a| < \epsilon$
What algebraic manipulations do we need to do to express $|\sqrt x - \sqrt a|$ as a multiple of $\delta$?
$|\sqrt x - \sqrt a| = |\frac {x-a}{\sqrt x + \sqrt a}|$
$\sqrt x + \sqrt a > \sqrt a$ 
If $a\ne 0$ we can say when $\delta < \sqrt a\epsilon, |\sqrt x - \sqrt a|<\epsilon.$
But what about when $a = 0$? In this case, a two-sided limit does not exist.
There is a neighborhood of the domain around $a$ that maps arbitrarily close to $\sqrt a$
A: Your reasoning does not seem correct (honestly, I really can't follow it). Here's what I have.
Proof. Let $\epsilon > 0.$ Choose $\delta = \epsilon |\sqrt{x}+\sqrt{a}|.$ Then when $|x - a| < \delta,$
\begin{align*}
|\sqrt{x} - \sqrt{a} | &= \frac{x-a}{|\sqrt{x}+\sqrt{a}|} \\
&< \frac{\delta}{|\sqrt{x} + \sqrt{a} |} \\
&< \epsilon
\end{align*}
as required.
I welcome any refinements.
