# 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.

• Did you try to solve this question? – TZakrevskiy Dec 13 '14 at 22:23
• what do we else know about $f(x)$? – Dr. Sonnhard Graubner Dec 13 '14 at 22:23
• @Dr.SonnhardGraubner Why do we need anything else? – TZakrevskiy Dec 13 '14 at 22:24
• try to use the rules of De L'Hospital – Dr. Sonnhard Graubner Dec 13 '14 at 22:29
• @TZakrevskiy "Dr." never answers to comments, especially those raising legitimate concerns. – Did Dec 14 '14 at 16:08

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 $$\lim_{x\to0}\frac{f(x)-\sin{f(x)}}{x^3}=\frac{(f'(0))^3}{6}.$$

• $f(x)=ax+O(x^2)$, since we know that $f(0)=0$, not $f'(0)=0$. – Jack D'Aurizio Dec 13 '14 at 22:31
• @JackD'Aurizio Oops... Thank you very much Jack! – Olivier Oloa Dec 13 '14 at 22:36
• Can you really conclude that $f(x)=f'(0)x+O(x^2)$ without using the second derivative? What about $f(x) = |x|^{3/2}$ ? – Martin R Dec 13 '14 at 22:46
• It seems that in both answers we are using that a second derivative of f exists. Otherwise how do we get the remainder term in the Taylor expansion? – 1234 Dec 13 '14 at 22:57
• @1234 I've edited my answer, to get a correct one, taking into account your remark, even if the conclusion remains the same. Thank you. – Olivier Oloa Dec 13 '14 at 23:40

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 $$\frac{f(x)-\sin f(x)}{x^3} = \frac{f^\prime(0)^3}{6} + o(1) \ .$$

• Where does the remainder term in the Taylor expansion of f(x) come from? We only have that f is once differentiable. – 1234 Dec 13 '14 at 23:00
• This is just writing $\lim \frac{f(x)}{x} = f^\prime(0)$. (the "error" term goes to 0, i.e. is o(1)) – Clement C. Dec 13 '14 at 23:03
• But what is the actual error term here? The next term of the Taylor expansion would be $\frac{1}{2}f''(\psi)x^2$ but we cannot write this without the existence of a second derivative. – 1234 Dec 13 '14 at 23:07
• you don't need to know the error term to compute the limit, as long as the error is of lower order in x than the limit itself (that is, as long as the limit is defined). What's the leading order error is irrelevant – Ferdinando Randisi Dec 14 '14 at 0:25

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.