calculus - Derivative by definition

I'm trying to find the derivative by definition of the following function:

$f(x)=\sqrt{|x|}\sin(x)$

I know that by definition:

$$f'(x)=\lim_{h\to0}\frac{\sqrt{|x+h|}\sin(x+h)-\sqrt{|x|}\sin(x)}{h}$$

But if I try to find the derivative at $0$ I get:

$$f'(0)=\lim_{h\to0}\frac{\sqrt{|h|}\sin(h)}{h}=0$$

which is not true because the derivative $DNE$ at $0$
because:

$$f'(x)=\begin{cases} \sqrt{x} \cos x+\frac{\sin x}{2\sqrt{x}}, \ \ \ \ \ x>0 \\ \sqrt{-x} \cos x-\frac{\sin x}{2\sqrt{-x}}\ \ \ x<0 \\ \end{cases}$$

So how can it be that the derivative exists only when it is calculated by definition?

Answer

You shouldn't say that the derivative does not exist.

Indeed,

$$\lim_{h\to0}\frac{\sqrt{|h|}\sin(h)}{h}=\lim_{h\to0}\sqrt{|h|}\cdot\lim_{h\to0}\frac{\sin(h)}{h}=0\cdot1.$$

As the limit exists, this is the value of the derivative.