TITLE: How to calculate $\lim_{x \to0} \dfrac{f(x)-f(\ln(1+x))}{x^{3}}$
QUESTION [6 upvotes]: $f$ 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.

REPLY [2 votes]: 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:
$$L = \frac{3f^{\prime\prime}(0)-2f^\prime(0)}{6}=\frac{12}{6}=2.$$