Real Analysis Question concerning existence of curve and derivative? So I'm doing some self study on Real Analysis from a friend's set of notes and classwork from last semester.  I'm trying to prepare myself for next year.  Anyway, I got caught up in the following question and was looking for some direction.  With out explicitly knowing what $f(x)$ is how do I go about showing that $f(0)$, $f'(0)$ exist?  Do I have to choose a function? 
Suppose that $f \colon R \to R$ is differentiable everywhere, and that
$$\lim_{x \to 0} \frac{f(x)}{x^2} = L$$
for some real number $L$.


*

*Find $f(0)$. Prove your answer.

*item Find $f'(0)$. Prove your answer.

*Give an example of a function $f$ satisfying the above conditions such that $f''(0)$ does not exist. Prove your answer.

 A: The claim that $f$ is everywhere differentiable actually tells you everything you need: it tells you that $f$ exists everywhere (since the definition of the derivative doesn't work if the function doesn't exist),it tells you that $f'$ exists everywhere, and it tells you that $f$ is continuous everywhere (since differentiability implies continuity).
To address the questions asked:


*

*We know that $x^2\to0$ as $x\to0$. As such, $f(x)$ can only have one limit as $x\to0$ to prevent the $\frac{1}{x^2}$ from blowing the whole thing up to infinity.  What is that limit?  Then, use continuity to deduce $f(0)$ from the limit of $f(x)$ as $x\to0$.

*Remember that
$$
f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}.
$$
You know $f(0)$ from $1$; plug this in to get a limit expression for $f'(0)$. This statement will be related to the limit given in the assumptions, but not equal to it; try to think about what the limit must be to make the limit in the assumptions possible.
A: Hint: For the example, try $f(x)=x^3 \sin(1/x)$ when $x\ne 0$, and $f(0)=0$. 
