Inverse rule to the L'Hôpital's rule If in L'Hôpital's rule we have that: $f,g :(a,b)\to \mathbb{R}$, there exist $f'(x)$, $g'(x)$, and $g'(x)\ne0$, 
$$\lim_{x\to a^+} \frac{f(x)}{g(x)} = L,$$
and also $\lim_{x\to a^+} f(x)=\lim_{x\to a^+} g(x) = 0$, must it be also
$$\lim_{x\to a^+} \frac{f'(x)}{g'(x)} = L$$
or not? I think it is wrong in some cases, but I can't find an example.
 A: The exception is that there may be a case that $\lim_{x\to a^+} f'(x)$ does not exist.
So let's define 
$$g(x)=x, \quad f(x)=\int_0^x(1+\sin(1/y))dy$$. 
Apparently $\lim_{x\to 0^+}g(x)=\lim_{x\to 0^+}f(x)=0$, and $\lim_{x\to 0^+} f'(x)$ does not exist.
The tricky part is to show that $\lim_{x\to 0^+} f(x)/g(x)=1$. But first of all, it is true intuitively because $\sin(1/y)$ oscillates infinitely fast so the integral cancels out.
More rigorously, we can treat the integral by define $z=1/y$
$$
\int_0^x\sin(1/y)dy=\int_{1/x}^\infty \frac{\sin(z)}{z^2}dz
$$
It can be shown that $\int_\alpha^\infty[\sin(z)/z^2]dz$ is bounded by two sums
$$
\sum_{n=N}^\infty a_n>\int_\alpha^\infty\frac{\sin(z)}{z^2}dz>\sum_{n=N}^\infty b_n
$$
Where $N\equiv \lfloor \alpha/(2\pi)\rfloor$, $a_n$ and $b_n$ are defiend as 
$$
a_n\equiv \int_{2n\pi}^{(2n+2)\pi}\frac{\sin(z)}{z^2}dz, \quad b_n\equiv \int_{(2n-1)\pi}^{(2n+1)\pi}\frac{\sin(z)}{z^2}dz 
$$
$$
a_n=\int_{2n\pi}^{(2n+1)\pi}\sin(z)\left(\frac{1}{z^2}-\frac{1}{(z+\pi)^2}\right)dz
$$
So 
$$
0<a_n<\int_{2n\pi}^{(2n+1)\pi}\left(\frac{1}{z^2}-\frac{1}{(z+\pi)^2}\right)dz =\frac{2\pi^2}{2n(2n+1)(2n+2)\pi^3}
$$
Thus $a_n\sim\mathcal{O}(1/n^3)$, and $\sum_{n=N}^\infty a_n \sim \mathcal{O}(1/N^2)=\mathcal{O}(1/\alpha^2)$
btw,the properties of the sum can be found on wolframalpha if you are interested.
link 1 link 2
Similarly we can prove that $0>\sum_{n=N}^\infty b_n \sim \mathcal{O}(1/\alpha^2)$
So 
$$
\lim_{x\to0^+}\frac{1}{x}\int_0^x\sin(1/y)dy=\lim_{x\to0^+}\frac{1}{x}\int_{1/x}^\infty \frac{\sin(z)}{z^2}dz =\lim_{x\to0^+}\frac{1}{ x}\mathcal{O}(x^2)=0
$$
