Why is the derivative of $\frac{|x|}{x}$ equal to $\emptyset$ at $x=0$? I got a bit of a confusion here. If $\varphi(x)=\frac{|x|}{x}$, then 
$$
\varphi(x) = \left.\Bigg\{
\begin{array}{cc}
1 &if \ x>0\\
\emptyset & if \ x=0\\
-1 & if \ x <0
\end{array} \right.
$$
and then
$$
\varphi'(x) = \left.\Bigg\{
\begin{array}{cc}
0 &if \ x>0 \cup x<0\\
\emptyset & if \ x=0
\end{array} \right.
$$
The derivative at $0$ is nevertheless also $0$. Why? My only suspicion is that $\varphi(x)$ is actually somehow defined at $x=0$, although I do not see how: there should be an indeterminate form $\frac{0}{0}$. Some form of L'Hospital's rule? 
 A: The function $\phi(x) = |x|/x$ has a step discontinuity at $x=0$.  Not only is it discontinuous, but you cannot redefine $\phi(0)$ to make it continuous, let alone differentiable.
Some other motivational examples:
If your function were something like $\psi(x) = \sin x/x $, you would similarly have $\psi(0)$ undefined.  However, if you redefined $\psi(0) = 1$, you would have a function that is both continuous and differentiable at $x=0$. 
For $\gamma(x) = x^2/|x|$, you could set $\gamma(0) = 0$ to make the function continuous at $0$, but it still wouldn't be differentiable.
A: This is a very nice example. Clearly $|x|/x$ has problems when $x=0$. The function $\varphi$ is not what we call differentiable at $x=0$, even though it has a continuous derivative at $x=0$. A function can't be called differentiable unless it is itself continuous. Let's consider $x<0$ and $x>0$ separately. 
Assume that $x<0$. Let $x=-k$ where $k>0$. We have:
$$\frac{|x|}{x} = \frac{|-k|}{-k}=\frac{k}{-k}=-1 \, . $$
It follows that $\varphi(x)=-1$ for all $x<0$. Thus $\varphi$ is constant for all $x<0$ and so $\varphi'(x)=0$. It follows that $\varphi'(x)$ tends towards zero as $x$ tends towards zero from the negative side.
Assume that $x>0$. Let $x=k$ where $k>0$. We have:
$$\frac{|x|}{x} = \frac{|k|}{k}=\frac{k}{k}=1 \, . $$
It follows that $\varphi(x)=1$ for all $x>0$. Thus $\varphi$ is constant for all $x>0$ and so $\varphi'(x)=0$. It follows that $\varphi'(x)$ tends towards zero as $x$ tends towards zero from the positive side.
Putting these two answers together, we may conclude that $\varphi'$ is continuous because:
$$\lim_{x \to 0^-} \left( \frac{d}{dx}\frac{|x|}{x}\right) = \lim_{x \to 0^+} \left(\frac{d}{dx}\frac{|x|}{x}\right) = 0 \, . $$ 
However, $\varphi$ fails to be itself continuous because $\varphi(x)$ tends towards $-1$ as $x$ tends towards zero from the negative side while $\varphi(x)$ tends towards $+1$ as $x$ tends towards zero from the positive side, meaning that
$$\lim_{x \to 0^-} \frac{|x|}{x} \neq \lim_{x \to 0^+} \frac{|x|}{x} \, . $$
