Prove that $|x|$ isn't differentiable at 0 through epsilon-delta logic? I remember in high school a proof of this was to show that the right-hand and left-hand limits were not equal to each other. But I was wondering, coming off of learning the $\epsilon - \delta $ notation recently, if a more elegant method existed? I'm still fuzzy on the intricacies of this notation, so I was wondering if this example would help clarify it for me.

In fact, the function $x \mapsto |x|: \mathbb{R} \to \mathbb{R}$ is not differentiable at and only at $x := 0$. If $x < 0$ then $|x| = -x$, and the function $x \mapsto -x$ is differentiable; for $x > 0$ likewise.
But
$$
\frac{|0+h| - |0|}{h} = 1
$$
for all $h > 0$ and
$$
\frac{|0+h| - |0|}{h} = -1
$$
for all $h < 0$,
so the limit of the difference quotient as $h \to 0$ does not exist, and hence the function $x \mapsto |x|$ is not differentiable at $0$.
Taking any $0 < \varepsilon < 1$ suffices if you want epsilon-analysis.
