Question about $f :\mathbb{R}\rightarrow \mathbb{R}$ defined as $f(x)=|x|^{\frac{3}{2}}$ (TIFR GS $2010$) Question is :
 
I am not sure how to check for differentiability,
only thing i know is how to see for differentiability of $f(x)=|x|$
$\lim_ {x\rightarrow 0} \frac{f(x)}{x}=\lim_ {x\rightarrow 0} \frac{|x|}{x}$
But then, when $x$ approach $0$ from positive line we have $\lim_ {x\rightarrow 0^+} \frac{|x|}{x}=1$ and  when $x$ approach $0$ from negativeline we have $\lim_ {x\rightarrow 0^-} \frac{|x|}{x}=-1$ 
So, limit does not exists so the function $f(x)=|x|$ is not differentiable at $0$
But, I am not getting any idea how to proceed in case of $f(x)=|x|^{\frac{3}{2}}$ 
$lim_ {x\rightarrow 0} \frac{f(x)}{x}=\lim_ {x\rightarrow 0} \frac{|x|^{\frac{3}{2}}}{x}$ I was thinking to write this as 
$$\lim_ {x\rightarrow 0} \frac{|x|^{\frac{3}{2}}}{x}=\lim_ {x\rightarrow 0} \frac{|x|^{2-\frac{1}{2}}}{x}=\lim_ {x\rightarrow 0} \frac{|x|^2 .|x|^{\frac{-1}{2}}}{x}=\lim_ {x\rightarrow 0} \frac{x}{|x|^{\frac{1}{2}}}$$
as $x$ is in square rot in denominator i am unable to see what are its left and right limits for $x=0$
Please help me to proceed further..
Thank you
 A: $\newcommand{\+}{^{\dagger}}%
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$$
\totald{\pars{\verts{x}^{3/2}}}{x} = {3 \over 2}\,\verts{x}^{1/2}\sgn\pars{x}
$$
A: It might be easier to denote $|x|^{3/2} = x\cdot\sigma(x)\sqrt{|x|}$, where $\sigma(x)$ is the sign of $x$. Then
$$\left|\frac{|x|^{3/2}}{x}-0\right|=\sqrt{|x|}\to 0.$$
