This feels stupid enough, yet it escapes me. Someone help me?
Let $$f(x):=\begin{cases}x, & x<0 \\ x+1, & x\ge 0 \end{cases}.$$ Then since it is discontinuous at $x=0,$ one should not expect it to be differentiable there. To show this from the definition is what seems to be a problem for me, for we have $$\frac{f(0+\delta)-f(0)}{\delta}=\frac{(0+1+\delta)-(0+1)}{\delta}=1,$$ so that taking limits from either direction leaves it unfazed and solidly $1,$ which is nonsense since it cannot be differentiable there.
Please help!
$$\frac{f(0-\delta)-f(0)}{-\delta}=-\frac{0-\delta-1}\delta=1+\frac1\delta.$$
The right derivative exists, not the left one.