Show that $f(x) = \sqrt{x}$ is continous at any $a > 0$ We first note that if $|x-a| < \delta$ and $\delta \leq 3a$, we have: 
$|x-a|<3a \Leftrightarrow -3a < x-a < 3a \Leftrightarrow -2a < x < 4a \Leftrightarrow |x| < 4a \Leftrightarrow |\sqrt{x}| < 2\sqrt{a} $.
Let's also note that $|2\sqrt{a}-\sqrt{a}| \leq |\sqrt{x}-\sqrt{a}| $ and so, $ \frac{1}{|\sqrt{x}-\sqrt{a}|} \leq \frac{1}{|2\sqrt{a}-\sqrt{a}|} $.
Then, for all $ \varepsilon > 0 $ there exists $\delta = \min\{ 3a, \sqrt{a}\varepsilon \}$, if $|x-a|<\delta$:
\begin{align}
|f(x) - f(a)| &= |\sqrt{x} - \sqrt{a}| \\
&= \frac{|x-a|}{|\sqrt{x} - \sqrt{a}|} \\
&\leq \frac{|x-a|}{|2\sqrt{a}-\sqrt{a}|} \\
&=\frac{|x-a|}{\sqrt{a}} \\
&< \frac{\delta}{\sqrt{a}} \leq \frac{\sqrt{a}}{\sqrt{a}}\varepsilon = \varepsilon
\qquad \text{Q.E.D.}
\end{align} 

Is this proof correct?


If $x > 0$
and $h > 0$
then
$\begin{array}\\
|\sqrt{x+h}-\sqrt{x}|
&=|(\sqrt{x+h}-\sqrt{x})\dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}|\\
&=|\dfrac{h}{\sqrt{x+h}+\sqrt{x}}|\\
&\lt|\dfrac{h}{2\sqrt{x}}|\\
\end{array}
$
Therefore,
if
$h < 2\epsilon \sqrt{x}$
then
$|\sqrt{x+h}-\sqrt{x}|
\lt \epsilon$.
