Proving multivariable non-differentiability using this particular definition So I've been asked to prove non-differentiability using this particular method, and I'm a bit lost. I'm supposed to prove that $f(x,y) = |x-y|$ is not differentiable along the $x=y$ line. To do it, I'm supposed to use this definition of differentiability:
A function $f$ is called differentiable at a point $a$ in it's domain if there exists a vector $c\in\mathbb{R^{n}}$ such that:
$$\lim\limits_{\bf h \to \bf 0} \frac{f(\boldsymbol a + \boldsymbol h) - f(\boldsymbol a) - \boldsymbol c \cdot \boldsymbol h}{|h|} = 0$$
Okay, so obviously $c$ would be the gradient of $f$ at that point, and obviously it won't exist because there is an apex there, but I can't for the life of me prove that it doesn't exist. Any help would be appreciated.
 A: Let $x \in \mathbb{R}$, $h>0$, then
$$\left|\frac{f((x,x)^T+h \cdot e^1)-f((x,x)^T)-1 \cdot h}{|h|}  \right| = \left| \frac{|(h,0)^T|-1 \cdot h}{h} \right| \stackrel{h>0}{=} 0$$
On the other hand
$$\left|\frac{f((x,x)^T+h \cdot e^1)-f((x,x)^T)+1 \cdot h}{|h|} \right| = \left| \frac{|(h,0)^T|+1 \cdot h}{-h} \right| \stackrel{h<0}{=} 0$$
for $h<0$ where $e^1:=(1,0)^T$. This means that $f$ is not differentiable at $(x,x)$ (...since the previous equations show that $\partial_x f$ does not exist. Or if you want to prove it straight from the definition of differentiability: The first equation equation shows $c \cdot (h \cdot e^1) =c_1 \cdot h \stackrel{!}{=} 1 \cdot h$ whereas the second one shows $c \cdot (h \cdot e^1) = c_1 \cdot h \stackrel{!}{=} -1 \cdot h$ - and there exists obviously no such $c=(c_1,c_2) \in \mathbb{R}^2$.) 
Similar proof works for $n \geq 3$.
A: Consider the point ${\bf a}=(a,a)$ for some $a\in{\mathbb R}$. Assume that $f$ is differentiable at ${\bf a}$ according  to the above definition. As $f({\bf a})=0$ we would then simultaneously have
$$\lim_{{\bf h}\to{\bf 0}}{f({\bf a}+{\bf h})-{\bf c}\cdot {\bf h}\over |{\bf h}|}=0\ ,\quad
\lim_{{\bf h}\to{\bf 0}}{f({\bf a}-{\bf h})-{\bf c}\cdot(-{\bf h})\over |{\bf h}|}=0\ ,$$
and therefore 
$$\lim_{{\bf h}\to{\bf 0}}{f({\bf a}+{\bf h})+f({\bf a}-{\bf h})\over |{\bf h}|}=0\ .\qquad(*)$$
For real $h>0$ put ${\bf h}:=(h,-h)$. Then $|{\bf h}|=\sqrt{2} h$ and
$$f({\bf a}+{\bf h})+f({\bf a}-{\bf h})=\bigl|(a+h)-(a-h)\bigr|+\bigl|(a-h)-(a+h)\bigr|=4h\ .$$
As
$$\lim_{h\to 0+}{4h \over\sqrt{2} h}=2\sqrt{2}\ne0\ ,$$
condition $(*)$ is not fulfilled at this point ${\bf a}$.
