Does this limit really exist? I'm working with this question in an exam:
Suppose $f(x)$ is $C^1$ in some neighbourhood about $x=0$, and $f'(0)=0$, $f''(0)=1$
Then I was asked to calculate the limit
$$\lim_{x\to 0}{f(x)-f(\ln(1+x))\over x^3}$$
I'm confused about whether the conditions given was enough to find this limit, since by the Taylor formula, we can only know the second-order behavior (i.e. remainder is $o(x^2)$).
Does the limit really exist?
 A: Using only the given assumptions, without additional differentiability,
$$
\begin{align}
   \frac{f(x)-f(\ln(1+x))}{x^{3}} & =\frac{\int_{\ln(1+x)}^{x}f'(u)\,du}{x^{3}} \\
    & = \frac{1}{x^{3}}\int_{\ln(1+x)}^{x}\{f'(u)-f'(0)\}du+f'(0)\frac{x-\ln(1+x)}{x^{3}} \\
    & = \frac{1}{x^{3}}\int_{\ln(1+x)}^{x}\{f'(u)-f'(0)\}du \\
    & = \frac{1}{x^{3}}\int_{\ln(1+x)}^{x}\left[\frac{f'(u)-f'(0)}{u-0}-f''(0)\right]u\,du + f''(0)\frac{x^{2}-(\ln(1+x))^{2}}{2x^{3}}
\end{align}
$$
The second term on the right has a limit of $f''(0)/2$ as $x\rightarrow 0$ because
$$
\begin{align}
    \lim_{x\rightarrow 0} \frac{x^{2}-(\ln(1+x))^{2}}{x^{3}} & =
    \lim_{x\rightarrow 0} \frac{x-\ln(1+x)}{x^{2}}\lim_{x\rightarrow 0}\frac{x+\ln(1+x)}{x} \\
   & = \lim_{x\rightarrow 0}\frac{1-1/(1+x)}{2x}
       \lim_{x\rightarrow 0}\frac{1+1/(1+x)}{1} \\
   & = \lim_{x\rightarrow 0}\frac{x}{(1+x)2x}\cdot 2 = 1.
\end{align}
$$
For any $\epsilon > 0$, the integral term on the right above can be bounded by $\epsilon$
by choosing $0 < |x| < \delta$ because the bracketed expression tends to 0 as
$u\rightarrow 0$ by the assumption that $f''(0)$ exists, and because
$$
    \frac{1}{x^{3}}\int_{\ln(1+x)}^{x}u\,du = \frac{x^{2}-(\ln(1+x))^{2}}{2x^{3}}
$$
has already been been show to have a limit of $1/2$. Putting the pieces together gives
$$
   \lim_{x\rightarrow 0}\frac{f(x)-f(\ln(1+x))}{x^{3}}=\frac{f''(0)}{2}.
$$
Check the constants!
A: One way to approach such problems, (though not the most rigorous way is to let
$$
f(x) = 0+0 ~x + 1/2 x^2 + c x^3 
$$
where $c$ is just a place holder. What you have to show is that the limit does not depend on $c$ (Making the argument rigorous is not too difficult but you buy the rigor at the expense of insight).
Now apply L'Hospital theorem three times. The third derivative of the numerator is
$$
-\frac{6\,c\,{{\log}\left( x+1\right) }^{2}-18\,c\,{\log}\left( x+1\right) +2\,{\log}\left( x+1\right) -6\,c\,{x}^{3}-18\,c\,{x}^{2}-18\,c\,x-3}{{\left( x+1\right) }^{3}} 
$$
Setting $x=0$ we get the numerator = 3
Differentiating the denominator 3 times you get 6.
Hence the limit is $1/2$.
P.S: The denominator $(1+x)$ is benign as it is $1$ at $x=0$ and can be dropped from the first derivative (which is what I did first). However, I used maxima to work out all the derivatives so you can check your work against mine.
A: This is the complete worked out solution:
The $f$ has the Taylor's series
$$
f(x) = \frac{1}{2} x^2 + h(x)
$$
where $h$ is the third and higher order terms.
One way to keep track of $h$ is to replace it by $c x^3$ and revisit this only if the final answer depends on $c$.
The numerator is
$$
n(x) = \frac{x^2}{2} - \frac{log(x+1)^2}{2} + c x^3 - c \log(x+1)^3
$$
differentiating
$$
\begin{align}
\frac{d n(x)}{d x} &= x - \frac{\log(x+1)}{x+1} + 3 ~c ~x^2 - 3 ~c ~{\log(x+1)^2}{x+1} \\
&= \frac{x+x^2 -\log(x+1)- 3~c~\log(x+1)^2-log(x+1)+3~c~x^3+3~c~x^2}{x+1}
\end{align}
$$
We can drop $x+1$ since it is well behaved at $x=0$ and is one at $x=0$.
Differentiating once more and dropping $x+1$, we get
$$
\frac{(x+1) n'(x)}{dx} = 1+2~x-\frac{1}{x+1} +9~c~x^2+ 6~c~x  -\frac{6~c~\log(x+1)}{(x+1)}
$$
we can take $(x+1)$ as the common denominator to 
get
$$
(x+1)(1+2~x)-1 + \hbox{term involving $c$}
$$
[Sorry: Algebra is a mess and really will go to zero at the next step. Trust me :) ]
Differentiate once more and set $x=0$ to get $3$.
The key point I wish to make: Replace higher order terms by a simple monomial. If the final answer does not depend on the monomial you can always work out the details without this trick. 
A: I do not know if what I did is stupid or not and I should appreciate opinions.  
I just expanded separately as Taylor series (around x=0) f[x] and f[Log(1+x)] up to the third order. I did not use any assumptions and {f[x] - f[Log(1+x)]} just write    
x^2 f'[0] / 2 + x^3 (f''[0] / 2 - f'[0] / 3) + ...  
Inserting the condition f'[0] = 0 gives for the limit f''[0] / 2
