This question is pretty old, but based on its number of views, it probably deserves a more robust answer. In order to show that this limit exists, we must show that the left-handed limit is equal to the right-handed limit. But before we do that, let's make the following observation:$$\frac{\sqrt{|x|}}{x} = \begin{cases} \frac{1}{\sqrt{x}}, \quad &\text{if } x>0\\ -\frac{1}{\sqrt{-x}}, \quad &\text{if } x<0\end{cases}$$This observation arises from the fact that the numerator is always positive, but the denominator is the same sign as $x$.
With this in mind, we calculate the left and right-handed limits.
$$\text{LEFT:} \lim_{x \to 0^-}\frac{\sqrt{|x|}-\sqrt{|0|}}{x-0} = \lim_{x \to 0^-} \frac{\sqrt{|x|}}{x} = \lim_{x \to 0^-} -\frac{1}{\sqrt{-x}}=-\frac{1}{\sqrt{\text{small pos. number}}} = -\infty$$
$$\text{RIGHT:} \lim_{x \to 0^+}\frac{\sqrt{|x|}-\sqrt{|0|}}{x-0} = \lim_{x \to 0^+} \frac{\sqrt{|x|}}{x} = \lim_{x \to 0^+} \frac{1}{\sqrt{x}}=\frac{1}{\sqrt{\text{small pos. number}}} = +\infty$$
Since the left-handed limit and the right-handed limit are not the same, the limit does not exist, and therefore, the function is not differentiable at $x=0$.
