Calculating $f(0), f'(0), f''(0)$ for a function $f$ satisfying $\lim_{x \to 0} (1 + x + f(x)/x)^{1/x} = e^3$. I'm trying to do the following problem and could use some help(from Apostol, Calculus, Volume I, 7.11 Ex. 33 p. 291):

A funtion $f$ has a continuous third derivative everywhere and
  satisfies the relation
$$ \lim_{x \to 0} \left(1 + x + \dfrac{f(x)}{x}\right)^{1/x} = e^3.$$
Compute $f(0), f'(0), f''(0),$ and $\lim_{x \to 0} \left(1 + \frac{f(x)}{x}\right)^{1/x}$.
[Hint: If $\lim_{x \to 0} g(x) = A$, then $g(x) = A + o(1)$ as $x \to 0$.]

The book gives the following as answers: $f(0) = 0, f'(0) = 0, f''(0) = 4$ and the limit $= e^2$.
I cannot seem to make any forward progress on this.  It is the last Exercise in a section of exercises on taking limits by using polynomial expansions of functions.  (This set of Exercises is immediately before the section on L'Hopital's rule... I don't know if that is applicable here, but a solution without it is appreciated since Apostol intends this to be done without it.)
My initial attempts involve writing:
$$\begin{align*}
& \lim_{x \to 0} \left(1 + x + \dfrac{f(x)}{x}\right)^{1/x} &= e^3.\\
\implies & \lim_{x \to 0} \left(e^{(1/x)\log(1 + x + f(x)/x)}\right) &= e^3.\\
\implies & \lim_{x \to 0} \left(e^{-\frac{\log x}{x} + \frac{1}{x} \log (x + x^2 + f(x))}\right) &= e^3.
\end{align*}$$
I wanted to do this to attempt to get to a point that I could write a polynomial expansion of $\log$ at $0$, but I can't seem to make any progress on that front.  Maybe there is a better way to simplify things?
Thanks for any help.  Full solutions or hints are equally welcome.
 A: The hypothesis is that $\left(1+x+x^{-1}f(x)\right)^{1/x}=\mathrm e^{3+o(1)}$, that is, $1+x+x^{-1}f(x)=\mathrm e^{3x+o(x)}$. Since $\mathrm e^{3x+o(x)}=1+3x+o(x)$, one gets $x^{-1}f(x)=2x+o(x)$.
Thus $f(x)=2x^2+o(x^2)$. 
Assume that  $f$ is twice differentiable at $0$. Then $f(x)=f(0)+f'(0)x+\frac12f''(0)x^2+o(x^2)$, by Taylor's theorem. By the uniqueness of the coefficients of the expansion of $f$ at $0$, one gets $f(0)=f'(0)=0$ and $f''(0)=4$.
Note that the fact that $f$ is twice differentiable at $0$ is a hypothesis, and certainly not a consequence of the expansion of $f$ at $0$. (On the other hand, the hypothesis that $f$ has a continuous third derivative everywhere is stronger than needed.)
Finally, $x^{-1}f(x)=o(1)$ hence $\log(1+x^{-1}f(x))=x^{-1}f(x)+o(x^{-1}f(x))=2x+o(x)$ and $x^{-1}\log(1+x^{-1}f(x))=2+o(1)$. 
Taking exponentials of both sides yields $\left(1+x^{-1}f(x)\right)^{1/x}=\mathrm e^{2+o(1)}$, which is equivalent to the statement that $\left(1+x^{-1}f(x)\right)^{1/x}\to\mathrm e^{2}$ when $x\to0$.
