Prove that the second derivative of $x^{4/3}$ does not exist at zero? 
Prove that the second derivative of $x^{4/3}$ does not exist at zero?

a/ As $f(x)= x^{4/3}$ we have $f'(x)=\frac{4}{3} x^{\frac{1}{3}}$ and $f''(x)=\frac{4}{9} x^{-\frac{2}{3}}$
Both $f$ and $f'$ are defined at zero but $f''$ is not defined at zero. Therefore the second derivative does not exist at 0
Would this be sufficient?
b/ how do you prove this using $f''(0)=\lim\limits_{h \rightarrow 0} \frac{f'(0+h)-f'(0)}{h}$
I came up with
$$f''(0) =\lim\limits_{h \rightarrow 0} \frac {\frac{4}{3} h^\frac{1}{3} } {h} =\lim\limits_{h \rightarrow 0} \frac{4}{3} h^{-\frac{2}{3}} $$
That is incorrect. I do not know what I am doing wrong here.
c/ since the second derivative does not exist, does this mean we cannot write a Taylor polynomial out the $f(x)=x^{4/3}$ or order 2? what would the justification be?
 A: Well, let us check, at first $f$'s differentiability at $0$.
$$f'(0)=\lim_{x\to0^+}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0^+}\frac{x^{\frac{4}{3}}-0}{x-0}=\lim_{x\to0^+}x^{\frac{1}{3}}=0$$
Now, let us check $f'$ differentiability at $0$:
$$f''(x)=\lim_{x\to0^+}\frac{f'(x)-f'(0)}{x-0}=\lim_{x\to0^+}\frac{\frac{4}{3}x^{\frac{1}{3}}}{x}=\lim_{x\to0^+}\frac{4}{3\sqrt[3]{x^2}}=+\infty$$
since $\lim\limits_{x\to0^+}\sqrt[3]{x^2}=0$ and $\sqrt[3]{x^2}\geq0$.
So, $f$'s second derivative does not exist at $0$, since it has not a finite value.
A step further, 

Let $p>0$ and let $f(x)=x^p$, $x\geq0$. Then $f^{(k)}(0)$ exists exactly for
  every $k=0,1,\dots,[p]$, where with $f^{(k)}(0)$ we note the $k$-th
  derivative of $f$ at $0$ and with $f^{(0)}(0)$ we note $f(0)$.

Proof: Let $k\in\mathbb{N}\cup\{0\}$. We have, for every $x>0$:
$$f^{(k)}(x)=\underbrace{p(p-1)\dots(p-k+1)}_{q_k(p)}x^{p-k}=q_k(p)x^{p-k}$$
Now, at $x_0=0$, we have:
$$f^{(k)}(0)=\lim_{x\to0^+}\frac{f^{(k-1)}(x)-f^{(k-1)}(0)}{x-0}$$
It is easy to see, as in our example above, that for every $k\leq[p]$ the abovementioned limit exists and is equal to $0$ and, hence, $f$'s $k$-th derivative does exist at $0$. Now, for $k=[p]+1>p$, we have:
$$f^{([p]+1)}(0)=\lim_{x\to0^+}\frac{q_{[p]}(p)x^{p-[p]}}{x}=q_{[p]}(p)\lim_{x\to0^+}x^{p-[p]-1}=+\infty$$
since $p-[p]-1<0$, and the proof is now complete.
A: Your work is good. We have
$$
\frac{f'(0+h)-f'(0)}{h}=\frac{\frac{4}{3}\sqrt[3]{h}}{h}
=\frac{4}{3\sqrt[3]{h^2}}
$$
which has infinite limit for $h\to0$. Hence $f'$ is not differentiable at $0$.
There is no Taylor expansion around $0$ of degree $2$: if there exist $a_0$, $a_1$ and $a_2$ such that
$$
\lim_{x\to0}\frac{f(x)-a_0-a_1x-a_2x^2}{x^3}=0
$$
then $f$ is twice differentiable at $0$.
