Derivative of absolute value using epsilon delta

Solution 1:

The trick here is that since $x\neq 0$ and $h\ll |x|$

you can select $\delta=\frac 12|x|$ so that $|h|<\delta\implies x+h$ and $x$ have the same sign

(Rem: $\frac x2<x+h<\frac{3x}2$ if $x>0$ and $\frac {3x}2<x+h<\frac{x}2$ if $x<0$)

This means there exists $\sigma=\pm 1$ such that $|x|=\sigma x$ and $|x+h|=\sigma(x+h)$.

The expression becomes $$\dfrac{|x+h|-|x|}{h}=\dfrac{\sigma(x+h)-\sigma x}{h}=\dfrac{\sigma h}h=\sigma\to \sigma=\dfrac{|x|}{x}=\dfrac x{|x|}$$

You don't even need an epsilon, the expression is constant as soon as $h$ gets sufficiently small (in fact we can even refine to $|h|<|x|$ instead of $\frac 12|x|$).

See https://www.desmos.com/calculator/x0b076shwg