Show that $f(x) = \sqrt{x}$ is continous at any $a > 0$

Question

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?

Answer 1

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 $|\sqrt{x+h}-\sqrt{x}| \lt \epsilon$.

Answer 2

$|\sqrt x| \le |x| < 4a$ is not true if $x < 1$.

And thus $|\sqrt{x}| \le 4a$ need not be true. If $x < 1$ then $x < \sqrt{x}$ so if $\frac x4 < a < \frac{\sqrt a}4$ this will not be true.