How to calculate $\lim_{x \to0} \dfrac{f(x)-f(\ln(1+x))}{x^{3}}$ $f$ is a differntiable function on $[-1,1]$ and doubly differentiable  on $x=0$ and $f^{'}(0)=0,f^{"}(0)=4$. 
How to calculate $$\lim_{x \to0} \dfrac{f(x)-f\big(\ln(1+x)\big)}{x^{3}}.
$$
I have tried the L'Hosptial's rule, but get trouble with the form of 
$$\lim_{x \to 0}f^{"}(x)-\dfrac {f^{"}\big(\ln(1+x)\big)}{(1+x)^2},$$
it may not exist.
 A: Answer. The value of the limit is $\,\,\dfrac{f''(0)}{2}=2$.
Hint. Mean Value Theorem provides that
$$
f(x)-f\big(\log (1+x)\big)=\big(x-\log(1+x)\big)\,f'\big(\xi(x)\big),
$$
for some $\xi(x)\in\big(\log(1+x),x\big)$. Then
$$
\frac{f(x)-f(\log (1+x))}{x^3}=\frac{x-\log(1+x)}{x^2}\cdot\frac{f'(\xi(x))}{\xi(x)}\cdot
\frac{\xi(x)}{x}.
$$
Then $\dfrac{x-\log(1+x)}{x^2}\to \dfrac{1}{2}$ (L'Hopital), $\dfrac{f'(\xi(x))}{\xi(x)}=\dfrac{f'(\xi(x))-f'(0)}{\xi(x)}\to f''(0)$, and
$\dfrac{\xi(x)}{x}\to 1$, since $\log(1+x)<\xi(x)<x$.
A: I think you can use l'Hôpital's rule, but be careful with the chain rule:
First step:
$$L = \lim_{x\rightarrow0}\frac{f(x)-f(\ln(1+x))}{x^3} = \lim_{x\rightarrow0}\frac{f^\prime(x)-\frac{f^\prime(\ln(1+x))}{1+x}}{3x^2}.$$
Second step:
$$L = \lim_{x\rightarrow0}\frac{f^{\prime\prime}(x) - \frac{f^{\prime\prime}(\ln(1+x))}{(1+x)^2}+\frac{f^\prime(\ln(1+x))}{(1+x)^2}}{6x}.$$
Third step:
$$L = \lim_{x\rightarrow0}\frac{f^{\prime\prime\prime}(x)-\frac{f^{\prime\prime\prime}(\ln(1+x))}{(1+x)^2}+\frac{2f^{\prime\prime}(\ln(1+x))}{(1+x)^3}+\frac{f^{\prime\prime}(\ln(1+x))}{(1+x^2)}-\frac{2f^\prime(\ln(1+x))}{(1+x^3)}}{6}$$
Hence:
$$L = \frac{3f^{\prime\prime}(0)-2f^\prime(0)}{6}=\frac{12}{6}=2.$$
A: Using MVT :
$$\lim_{x \to0} \dfrac{f(x)-f(ln(1+x))}{x^{3}} =\lim_{x \to 0} \dfrac{f(x)-f(0)+f(\ln 1)-f(ln(1+x))}{x^{3}} = \lim_{x \to 0} \dfrac{xf'(\theta_1 x)-x[f(\ln (1+\theta_2 x)]'}{x^{3}} $$
