Real Analysis - limits and differentiation Consider a real-valued function $h:\mathbb{R} \rightarrow \mathbb{R}$ with the following properties


*

*$\lim\limits_{x \to 0} \frac{h(x)}{x^{2}}=K$

*h is differentiable everywhere
a. Find $h(0), h^{'}(0)$
b. Find a function $g$ such that properties 1 and 2 hold and $h^{''}(0)$ does not exist.
-> I've attempted to solve this using the limit definition of the derivative for $h{'}(0)$ but I keep going around in circles. Can someone please help. 
 A: Hint: If $g(x) \to 0$ as $x \to 0$ and $f(x)$ does not converge to zero, then the quantity 
$$\frac{f(x)}{g(x)}$$
is not bounded in a neighborhood of zero.
A: By T. Bongers's hint above, if $\lim\limits_{x \to 0}h(x) \ne 0$, then it is impossible for $\lim\limits_{x \to 0} \dfrac{h(x)}{x^2}$ to be finite. So, $h(0) = 0$.
Note that $$K = \lim_{x\to 0} \frac{h(x)}{x^2} = \lim_{x \to 0} \frac{\frac{h(x)}{x}}{x} = \lim_{x \to 0} \frac{\frac{h(x)-h(0)}{x-0}}{x}.$$
By the same reasoning as before, $\lim\limits_{x\to 0}\dfrac{h(x)-h(0)}{x-0}$ must be $0$ if the last limit above is finite. So $h'(0)=0$.

How can we make $h''(0)$ not exist? Look at this for inspiration.
Let $$h(x) = \begin{cases} Kx^2 + x^3 \sin \frac 1 x & x \ne 0 \\ 0 & x=0\end{cases}.$$ Then $$\dfrac{h(x)}{x^2} = K+x\sin\frac 1 x \to K$$ as $x \to 0$.
Moreover, $$h'(x) =\begin{cases} 2Kx + x^2 \sin \frac 1 x - x \cos \frac 1 x & x \ne 0 \\ 0 & x = 0\end{cases}.$$
But
$$\frac{h'(x)-h'(0)}{x-0} = \frac{h'(x)}{x} = 2K + x \sin \frac 1 x - \cos \frac 1 x$$
does not have a limit as $x \to 0$.
A: If $K \neq 0$, it means that $h(x)=Kx^2+O(x^3)$, because otherwise the limit would be either $0$ or infinite. You can easily see that both $h(0)$ and $h'(0)$ are 0.  
