I need to show that the absolute value of x is not differentiable at x=0 using epsilon - delta. Thank you.
My thoughts: The absolute value at 0 is where we have a sharp cusp, so it's obviously not differentiable there because that's where the sided limits do not match. In terms of epsilon-delta, we need an epsilon that creates an instance where delta cannot be true to pose a contradiction. I assume that an epsilon of 1/2 would work nicely for this cause, but am unsure where to get the contradiction. If we plug in 1/2 for epsilon, 0 for x, and 1 for the other variable, c, then we end up with 1<delta and 1/2<Derivative.
Note that, by definition of a derivative, for $\lvert x \rvert$ to be differentiable at $0$ the limit $$\lim_{x\to 0} \frac{\lvert x \rvert - \lvert 0 \rvert}{x - 0} = \lim_{x\to 0} \frac{\lvert x \rvert}{x}$$ must exist. Now you can apply the $\varepsilon$-$\delta$ definition to show this limit in fact doesn’t. Can you take it from here?
(If you need a hint at any point, see this answer for a full solution.)
P.S. The expression above may be written as $$\forall \varepsilon >0, \exists \delta > 0 \text{ s.t. } 0<\lvert x - 0 \rvert<\delta \implies \left\lvert \frac{\lvert x \rvert}{x} - L\right\rvert < \varepsilon $$ using $\varepsilon$-$\delta$ notation, where $L$ is the supposed limit. Please refer to my answer linked above for a detailed proof of why such $L$ could not exist.