Prove $\lim_{x\to a}\sqrt{x}=\sqrt{a}$ by definition Stewart solution error Let $a&gt;0$. 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  &lt; x $? Seems that way to me by the way they choose delta.
My solution :
$$|x-a|&lt; a   \implies   \frac{1}{\sqrt{x} + \sqrt{a}} &lt; \frac{1}{\sqrt{a}}$$
and choose delta accordingly. Is that correct?
Thanks for the help.
You have to discuss the inequality $|\sqrt{x}-\sqrt{a}|&lt;\varepsilon$. 

Now take $\delta=\min(\varepsilon\sqrt{a},a)$ and assume $|x-a|&lt;\delta$. Since $\sqrt{x}+\sqrt{a}&gt;\sqrt{a}$, we have
$$
|\sqrt{x}-\sqrt{a}|=
\frac{|x-a|}{\sqrt{x}+\sqrt{a}}&lt;
\frac{|x-a|}{\sqrt{a}}&lt;\frac{\delta}{\sqrt{a}}\le\varepsilon
$$
Your reasoning does not seem correct (honestly, I really can't follow it). Here's what I have.

Proof. Let $\epsilon &gt; 0.$ Choose $\delta = \epsilon |\sqrt{x}+\sqrt{a}|.$ Then when $|x - a| &lt; \delta,$

\begin{align*}
|\sqrt{x} - \sqrt{a} | &amp;= \frac{x-a}{|\sqrt{x}+\sqrt{a}|} \\
&amp;&lt; \frac{\delta}{|\sqrt{x} + \sqrt{a} |} \\
&amp;&lt; \epsilon
\end{align*}

as required.

I welcome any refinements.