if $f'''(x)$ is continuous everywhere and $\lim_{x \to 0}(1+x+ \frac{f(x)}{x})^{1/x}=e^3$ Compute $f''(0)$ 
if $f'''(x)$ is continuous everywhere and $$\lim_{x \to 0}(1+x+ \frac{f(x)}{x})^{1/x}=e^3$$ Compute $f''(0)$

The limit equals to $$\begin{align} \lim_{x \to 0} \frac{\log(1+x+ \frac{f(x)}{x})}{x}-3=0.  \end{align}$$ 
From $$\frac{\log(1+x+ \frac{f(x)}{x})}{x}-3=o(1)$$ as $x \to 0$, I get $$1+x+\frac{f(x)}{x} = e^{3x+o(x)},$$ and  $$f(x)=x(e^{3x+o(x)}-x-1),\frac{f(x)}{x}=e^{3x+o(x)}-x-1$$ as $x \to 0$. So both $f(0)$ and $f'(0)$ are $0$. Approximating $e^{3x+o(x)}=1+3x+o(x)$ I get $$\begin{align} f(x) &= x(1+3x+o(x)-x-1) =2x^2+o(x^2). \end{align}$$
Now I try to use the definition of derivative to calculate the $f''(x)$ 
$$f''(x)=\lim_{x \to 0}\frac{f'(x)-f'(0)}{x}=\lim_{x \to 0} \frac{f'(x)}{x}$$
I'm not sure whether I can get $f'(x)$ by differentiating the approximation $2x^2+o(x^2)$ and how to differentiate $o(x^2)$.
 A: I will use the following formula for $f''(0)$. We have 
$$
f(x)=f(0)+f'(0)x+f''(0)\,\frac{x^2}2+o(x^3),
$$
$$
f(-x)=f(0)-f'(0)x+f''(0)\,\frac{x^2}2-o(x^3).
$$
Adding and solving for $f''(0)$, we get
$$\tag{1}
f''(0)=\frac{f(x)+f(-x)-2f(0)}{x^2}+o(x).
$$
Starting from your 
$$
1+x+\frac{f(x)}{x} = e^{3x+o(x)},
$$
we have
$$\tag{2}
x+x^2+f(x)=xe^{3x+o(x)}.
$$
Then, taking $x=0$, we get  $f(0)=0$;  and, using $(1)$,
\begin{align}
f''(0)&=\lim_{x\to0}\frac{f(x)+f(-x)-2f(0)}{x^2}=\lim_{x\to0}\frac{f(x)+f(-x)}{x^2}\\ \ \\
&=\lim_{x\to0}\frac{xe^{3x+o(x)}-x-x^2-xe^{-3x+o(-x)}+x-x^2}{x^2}\\ \ \\
&=-2+\lim_{x\to0}\frac{e^{3x+o(x)}-e^{-3x+o(-x)}}{x}\\ \ \\
&=-2+\lim_{x\to0}\frac{3x+3x+o(x^2)}{x}\\ \ \\ &=-2+6=4.
\end{align}
In the last limit I'm cancelling $o(x)$ with $o(-x)$; this is ok since both terms are coming from the same expression $(2)$.
A: Taking logs (as done in OP's post) we can see that $$\lim_{x \to 0}\dfrac{\log\left(1 + x + \dfrac{f(x)}{x}\right)}{x} = 3$$ or $$\lim_{x \to 0}\log\left(1 + x + \frac{f(x)}{x}\right) = 0$$ or $$\lim_{x \to 0}1 + x + \frac{f(x)}{x} = 1$$ or $$\lim_{x \to 0}\frac{f(x)}{x} = 0 \Rightarrow \lim_{x \to 0}f(x) = 0$$ and hence by continuity $f(0) = 0$. Now we can see that $$f'(0) = \lim_{x \to 0}\frac{f(x) - f(0)}{x} = \lim_{x \to 0}\frac{f(x)}{x} = 0$$ Now let $f''(0) = a$ and we have via Taylor's theorem $$f(x) = f(0) + xf'(0) + \frac{x^{2}}{2}f''(0) + o(x^{2})$$ This assumes only the existence of $f''(0)$ and nothing more. We have thus finally $$f(x) = \frac{ax^{2}}{2} + o(x^{2})$$ and therefore
\begin{align}
3 &= \lim_{x \to 0}\dfrac{\log\left(1 + x + \dfrac{f(x)}{x}\right)}{x}\notag\\
&= \lim_{x \to 0}\dfrac{\log\left(1 + x + \dfrac{ax}{2} + o(x)\right)}{x}\notag\\
&= \lim_{x \to 0}\dfrac{\log\left(1 + x + \dfrac{ax}{2} + o(x)\right)}{x + \dfrac{ax}{2} + o(x)}\cdot\dfrac{x + \dfrac{ax}{2} + o(x)}{x}\notag\\
&= \lim_{x \to 0}\left(1 + \frac{a}{2} + o(1)\right)\notag\\
&= 1 + \frac{a}{2}\notag
\end{align}
and hence $a = f''(0) = 4$.
