Question

Is the function ƒ(x)=|x^3| differentiable at x=0? Justify using limits.

Answer 1

Note that \(f(x)=|x^3|\) can be written as follows:
\[\Large f: \begin{cases} \mathbb{R} &\longrightarrow \mathbb{R} \\x & \longmapsto \begin{cases} x^3 &\forall x >0 \\ 0 & x=0\\ -x^3& \forall x < 0 \end{cases} \end{cases} \]
Which is the exact same function, just written differently, you can verify that for yourself. Clearly for all \(x \neq 0\) the function is differentiable, in special the function is differentiable at \(x=0\).

So what are is left to show for you? You need to make sure that the the function as also differentiable at \(x=0\) as \(x\) approaches \(0\). The way I have defined the above function we need to do this from the left and from the right hand side. I will do one case, you can do the other:
\[\large \lim_{h \to 0+} \frac{f(x=0+h)-f(x=0)}{h}=\lim_{h \to 0^+} \frac{f(h)-f(0)}{h}=\lim_{h\to 0^+} \frac{h^3}{h} \]
so you get: \[\large \lim_{h \to 0^+} h^2=0 \]
which shows that \(f\) is differentiable at \(x=0\) and \(f'(0)=0\)