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Is the square root of the absolute value function ($\sqrt{|x|}$) differentiable at $x = 0$?

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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 $x_0 = 0$ .


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