Proving that $\lim\limits_{h \to 0}\frac{|x+h|-|x|}{h}$ does not exist at $x=0$ In our exam, we are tasked to prove that $U(0)$ does not exist where,
$$U(x) = \lim\limits_{h \to 0}\frac{|x+h|-|x|}{h}$$

What I did was to simplify the function first, so I initially performed the limit operation. That function above is essentially the derivative of $|x|$, so I got,
$$U(x) = \frac{x}{|x|}$$
And after doing $U(0)$, it became
$$U(0) = undefined$$
This completed my proof. My teacher marked it wrong. What is the incorrect step in what I did and what could be the right proof?  
 A: You basically did two things in the opposite order as expected, and that allowed a fallacy to creep in.  You were asked first to substitute $x=0$, then evaluate the limit as $h\to 0$.  This is
$$
    U(0) = \lim_{h\to 0} \frac{|0+h| - |0|}{h} = \lim_{h\to 0} \frac{|h|}{h}
$$
which does not exist.
Instead, you took the limit as $h\to0$, (which does, as you say, compute the derivative at $x$), then asserted that this expression was not defined at $x=0$.  But this line of reasoning is not valid to show that the the original limit does not exist.  For instance, in your problem, replace $|x|$ by $$f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right) & \text{if $x\neq 0$} \\ 0 & \text{$x=0$}\end{cases}$$  Then if $x\neq0$
\begin{align*}
    f'(x) &= 2x \sin\left(\frac{1}{x}\right) + x^2 \cos\left(\frac{1}{x}\right)\cdot \left(-\frac{1}{x^2}\right)\\
    &= 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)
\end{align*}
Not only is this undefined at $0$, $\lim_{x\to 0} f'(x)$ does not exist.  But if you first set $x=0$, then take the limit, you get
$$
    \lim_{h\to 0} \frac{f(h) - 0}{h}
    = \lim_{h\to 0} h \sin\left(\frac{1}{h}\right)
    = 0
$$
So $f'(0) =0$.
A: You must strictly keep the definition given, so by what you say:
$$H(0)\stackrel?=\lim_{h\to0}\frac{|h|}h$$
and now show the above limit doesn't exist by calculating both one-sided limits
A: Your first step, that $U(x)=\frac {d|x|}{dx},$ is only true for $x\ne 0.$ So you obtain the result that $\lim_{x\to 0}U(x)$ does not exist. But you were not asked about $\lim_{x\to 0}U(x).$ You were asked about $U(0).$  Your first step says nothing about $U(0).$ You cannot assume that if  $\lim_{x\to 0}U(x)$  does not exist then $U(0)$ doesn't exist. 
