Find the limit: $\lim_{x\to0}\frac{f(x)-\sin{f(x)}}{x^3}$ If $f$ is differentiable at $0$ and $f(0)=0$ then show that the following limit exists and find it: $$\lim_{x\to0}\frac{f(x)-\sin{f(x)}}{x^3}$$
I have tried using L'Hospital's rule but without continuity of the first derivative we meet a dead end.  I have also tried using Taylor expansion but I don't think this works without the existence of a second derivative and we would also need differentiability on an interval, not just one point.
This is a preliminary exam practice question, so I must be able to prove it rigorously.
Here is one attempt (that I believe is flawed):
$$\lim_{x\to0}\frac{f(x)-\sin{f(x)}}{x^3}=\left(\lim_{x\to0}\frac{1}{x^2}\right)\left(\lim_{x\to0}\frac{f(x)}{x}-\lim_{x\to0}\frac{\sin{f(x)}}{x}\right)$$ $$=\lim_{x\to0}\frac{1}{x^2}\left(f'(0)-f'(0)\cos{f(0)}\right)=\lim_{x\to0}\frac{1}{x^2}*0=0$$
I don't think I am allowed to do it this way because the limit of each individual factor must exist in order to break the limit into a product of limits.
 A: For $y\to 0$, $\;$ $\sin y = y - \frac{y^3}{6} + o(y^3)$. 
Since $f$ is continuous at $0$ and $f(0)=0$, we get
$$
 \sin f(x) = f(x) - \frac{f(x)^3}{6} + o(f(x)^3),
$$
and thus
$$
\frac{f(x)-\sin f(x)}{x^3} = \frac{f(x)^3 + o(f(x)^3) }{6x^3}\ .
$$
Using the differentiability of $f$ at $0$, we also get that $\frac{f(x)}{x} = f^\prime(0) + o(1)$, and therefore that $$\frac{f^3(x)}{x^3} = f^\prime(0)^3 + o(1).$$ 
Plugging this back in the above, we get 
$$
\frac{f(x)-\sin f(x)}{x^3} = \frac{f^\prime(0)^3}{6} + o(1) \ .
$$
A: From our hypotheses, we deduce that near $0$,
$$
f(x)=f'(0)x+x\epsilon (x), \quad 
$$ with $\displaystyle \lim_{x\to0}\epsilon (x) =0$, then, since for $u$ near $0$,
$$
\sin u = u -\frac{u^3}{6}+\mathcal{o}(u^3)
$$
we have
$$
\sin f(x) = f(x)-\frac{(f'(0))^3}{6} x^3+x^3\epsilon_1 (x)
$$ with $\displaystyle \lim_{x\to0}\epsilon_1 (x) =0$ thus
$$\lim_{x\to0}\frac{f(x)-\sin{f(x)}}{x^3}=\frac{(f'(0))^3}{6}. $$
A: Here is an approach that's worth considering even though (or maybe because) it's not completely rigorous.
Because $f'(0)$ exists, we know that $f$ is continuous at $0$, hence $u=f(x)\to f(0)=0$ as $x\to0$.  This makes it tempting to write
$$\begin{align}
\lim_{x\to0}{f(x)-\sin f(x)\over x^3}&=\lim_{x\to0}\left({f(x)-\sin f(x)\over (f(x))^3}\cdot{(f(x))^3\over x^3}\right)\\
&=\left(\lim_{x\to0}{f(x)-\sin f(x)\over (f(x))^3} \right)\left(\lim_{x\to0}{f(x)\over x}\right)^3\\
&=\left(\lim_{u\to0}{u-\sin u\over u^3} \right)\left(\lim_{x\to0}{f(x)-f(0)\over x-0} \right)^3\\
&={1\over6}(f'(0))^3
\end{align}$$
using L'Hopital (or whatever you like) to evaluate the $u$-limit.
Where this falls short of rigor is in the fact that $f(x)$ may equal $0$ for infinitely many $x$ near $0$, so that the expression
$$\lim_{x\to0}{f(x)-\sin f(x)\over (f(x))^3}$$
may not be well defined.  It's possible to plug this gap with a separate argument to cover the problematic case, but probably not worth it, in light of the answers by Olivier Oloa and Clement C.  Mostly, it's worth taking note of the issues involved in "distributing" limits.  In particular, as the OP noted, when you go to evaluate a limit by breaking it into pieces, you have to take care that all the individual pieces exist.  
