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.