Is a $o(x^2)$ (Landau's notation) differentiable in a neighbourhood of 0?

Question

I wonder if a function $f:\textbf{R} \rightarrow \textbf{R}$ which has the next property: $$\lim_{x\rightarrow0}\frac{f(x)}{x^2}=0$$

is differentiable at an neighbourhood of $x=0$ (is trivially differentiable at x=0)

No further assumptions are made on $f$. I don't know where to start, although my intuition tells me that this is false.

Thanks

Answer 1

If $f(0)\ne0$, then $f$ is discontinuous at $x=0$. If $f(0)=0$, then $$ \frac{f(x)-f(0)}{x}=x\,\frac{f(x)}{x^2} $$ and $f'(0)=0$. But there is noting you can tell about differentiability at $x\ne0$. Consider for instance $$ f(x)=\begin{cases} x^3 & \text{if }x\in\Bbb Q,\\0 & \text{if }x\notin\Bbb Q. \end{cases} $$

Answer 2

For any $n\ge 1$, consider $f(x) = x^{n+1}\sin(x^{-(n+3)})$. Then $f(x) = o(x^n)$ as $x \to 0$.

But

$\begin{array}\\ f'(x) &=(n+1)x^{n}\sin(x^{-(n+3)})+x^{n+1}(-(n+3))x^{-(n+4)}(\cos(x^{-(n+3)})\\ &=(n+1)x^{n}\sin(x^{-(n+3)})-(n+3)x^{-3}\cos(x^{-(n+3)})\\ \end{array} $

and $\lim_{x \to 0} f'(x)$ does not exist.