For the function $f(x) = |x| + x$, how can one prove that it is not differentiable at the point $x = 0$? Specifically, I wish to use the Definition of the Derivative
$$\lim_{\Delta x \to 0}\frac{f(x + \Delta x) - f(x)}{\Delta x}$$
to attain two non-equal limiting values by taking $\Delta x$ to be first positive, and then negative.
It was my original thought to rewrite $f(x)$ as
$$ f(x) = \begin{cases} 2x&\text{if}\, x\geq 0\\ 0&\text{if}\, x < 0 \end{cases} $$
and then take the derivative for each separate case and show that they aren't equal, but I wish to use the above definition with both a positive and a negative $\Delta x$, and so the cases method would fall out of line with that.
Thank you.
If $\Delta x>0$, then $$\frac{f(\Delta x)-f(0)}{\Delta x}=2,$$whereas if $\Delta x<0$, then $$\frac{f(\Delta x)-f(0)}{\Delta x}=0.$$Therefore$$\lim_{\Delta x\to0^+}\frac{f(\Delta x)-f(0)}{\Delta x}=2\neq0=\lim_{\Delta x\to0^-}\frac{f(\Delta x)-f(0)}{\Delta x}.$$