Is the square root of the absolute value function ($\sqrt{|x|}$) differentiable at $x = 0$? I've been trying to solve the problem below for hours but so far I  haven't managed to find a solution. Help would really be appreciated. Thanks a lot!
Problem
Show whether or not the function $f(x)=\sqrt{|x|}$ is differentiable at $x_0 = 0$ by verifying if the limit  $\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}$ exists, that is, the limit as $x \to x_0$ from the left ($x \lt x_0$) is the same as as when approached from the right ($x \gt x_0)$ .
Own approach
Right limit: $$\lim_{x \to x_0} \frac{\sqrt{(x)}-\sqrt{({x_0})}}{x -x_0}=\frac{\sqrt{x}-\sqrt{x_0}}{x-x_0}=\frac{\sqrt{x}+\sqrt{0}}{x-0}=\frac{1}{\sqrt{x}}$$
Left limit: $$\lim_{x \to x_0} \frac{\sqrt{-1(-x)}-\sqrt{(x_0)}}{x-x_0}=\frac{\sqrt{x}-\sqrt{x_0}}{x-x_0}=\frac{\sqrt{x}+\sqrt{0}}{x-0}=\frac{1}{\sqrt{x}}$$
Left and right-hand limits are the same, the limit must therefore exist and $f(x)$ is thus differentiable at $x_0 = 0$.
Solution
According to the solution the limit does not exist, thus $f(x)$ not differentiable at $x_0 = 0$ .
 A: 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 $x=0$.
A: If you plot Graph of $f(x)=√|x|$ you will get the answer.
If you draw tangent to that graph at $x=0$ it will be a vertical tangent. Now also you have to understand that   $f'(x)=1/√|x|$ is not defined at $x=0$. So  function $f(x)=√|x|$ is not differential at $x_0=0$ but its continuous at $x_0=0$.
