If $f(x) = |x|^3$, show $f'''(0)$ does not exist If $f(x) = |x|^3$, show $f'''(0)$ does not exist.
So I've computed the first, second and third derivatives, so 
$f'(x) = 3x|x|$
$f''(x) = \begin{cases}
0,  & \text{if x = 0} \\
\frac{6x^2}{|x|}, & \text{if x $\ne$ 0}
\end{cases}$
$f'''(x) = \frac{6x^3}{|x|^3}$. 
I know if I were to simply plug  in $0$ for $x$, I would end up with $\frac{0}{0} = 0$, but that doesn't bring me to the $DNE$ conclusion. Is this someplace I should be using L'Hopital's Rule? The question didn't use the word "limit", but it's the only option I can come up with...
 A: Most straightforward is to say:
$f(x) = \begin{cases} -x^3 & x < 0\\x^3 & x\ge 0 \end{cases}$
The derivatives of $f$ will have the same piece-wise structure.
$f'''(x)$ will appear to have a jump discontinuity.  
The derivative is defined as a limit, if the left hand limit does not equal the right hand limit, the limit does not exist.  The derivative does not exist at the point of this jump.
A: By L'Hopital rule we have that
$$lim_{x\rightarrow 0^+}\frac{ f{''}(x)}{x}=12$$ and $$lim_{x\rightarrow 0^-}\frac{ f{''}(x)}{x}=-12$$ 
So the third derivative at $0$ does not exist.
A: You can also think of $f(x)=|x|^3$ as composite function $f(x)=(g\circ h)(x)$ with $g(x)=x^3$ and $h(x)=|x|$. By chain rule if at a point $x=c$ the function $h$ is differentiable and $g$ is differentiable at $h(c)$ then $f'(c)=g'(h(c))h'(c)$. But at $c=0$ we have that $h$ is not differentiable since $h'(0^-)=-1\neq 1=h'(0^+)$. As consequence $f$ is not differentiable at $c=0$. Surely then the third derivative doesn't exist there for it requires the first derivative to exist in the first place.
A: Let us first establish a lemma:

Given any function $g : (-\epsilon,\epsilon) \to \mathbb{R}$. If $g'(0)$ exists , then$\color{blue}{{}^{[1]}}$
  $$\mathcal{J}(g) \stackrel{def}{=} \lim_{h\to 0+} \frac{g(2h)-g(h) + g(-2h) - g(-h)}{h} = 0$$

Let's say $g'(0)$ exists and equals to $L$, we have
$$\lim_{h\to 0} \frac{g(2h)-g(h)}{h} = 
\lim_{h\to 0} \left[2\left(\frac{g(2h)-g(0)}{2h}\right) - \frac{g(h)-g(0)}{h}
\right] = 2L - L = L
$$
This implies
$$\mathcal{J}(g) = \lim_{h\to 0+} \left[\frac{g(2h)-g(h)}{h} - \frac{g(-2h)-g(-h)}{-h}\right] = L - L = 0$$
and justify the lemma.
Back to original problem. 
Since $f(x)$ is smooth for $x \ne 0$, it is easy to deduce
$f''(x) = 6|x|$ for $x \ne 0$. By direct computation, we have
$$\begin{align}\mathcal{J}(f'') &= \lim_{h\to 0+} \frac{f''(2h)-f''(h) + f''(-2h) - f''(-h)}{h}\\
&= \lim_{h\to 0+} \frac{6|2h| - 6|h| + 6|-2h| - 6|-h|}{h}\\
&= 12
\end{align}$$
Since $\mathcal{J}(f'') \ne 0$, by above lemma, $f'''(0)$ does not exist.
Notes


*

*$\color{blue}{[1]}$ - $\mathcal{J}(g)$ is the size of jump discontinuity of $g'(x)$ at $x = 0$. The whole point of this complicated mess is to have a formula for the jump in $f''(x)$ at $x = 0$ without explicit referral to values of $f(0)$, $f'(0)$, $f''(0)$ and $f'''(0)$ which are clumsy to describe and justify.

A: $$f(x)=\begin{cases} -x^3 & x < 0\\x^3 & x\ge 0 \end{cases}\Rightarrow f'(x)=\begin{cases} -3x^2=3x(-x) & x < 0\\3x^2=3x(x) & x\ge 0 \end{cases}\iff f'(x)=3x|x|$$
$$f^{''}(x)=\begin{cases} -6x & x < 0\\6x & x\ge 0 \end{cases}\iff f^{''}(x)=6|x|$$ It follows that the tangents are constantly $-6$ at the left of $0$ and constantly $6$ at the right of $0$ so $f^{''}(x)$ is not differentiable at $0$.
NOTE.-You can verify, for example, drawing the graphics that $6|x|=\begin{cases}
0,  & \text{if x = 0} \\
\dfrac{6x^2}{|x|}, & \text{if x $\ne$ 0}
\end{cases}$.
I mean your expression and mine are equivalent.
