# Can the quotient of derivatives be discontinuous?

Do there exist $$f,g$$ such that $$f(0)=0=g(0), g'(x)\neq 0$$ for all $$x,$$ and $$\lim_{x \to 0}\frac{f'(x)}{g'(x)}$$ exists, but $$\frac{f'(0)}{g'(0)} \neq \lim_{x \to 0}\frac{f'(x)}{g'(x)}?$$

My initial thought was to let $$f(x)=x^2\sin(1/x),$$ the classical example of a function with discontinuous derivative. However, this would require $$g$$ to be something like $$\int \cos(1/x)dx$$ which is not elementary.

• Actually, $\int\cos^2(x)\,\mathrm dx=\frac x2+\frac12\sin(x)\cos(x)$. Commented Apr 8 at 23:31
• Thanks for the corrections. @JoséCarlosSantos I did mean $\int \cos(1/x)dx$ Commented Apr 9 at 0:11

Since $$f$$ and $$g$$ are both differentiable at $$x=0$$, $$f(x) \rightarrow f(0)=0$$ and $$g(x) \rightarrow g(0)=0$$ as $$x\rightarrow 0$$. Therefore, you can use L'Hopital to see that $$\lim_{x\rightarrow 0} \frac{f(x)}{g(x)} = \lim_{x\rightarrow 0} \frac{f'(x)}{g'(x)}.$$ On the other hand, $$\lim_{x\rightarrow 0} \frac{f(x)}{g(x)} = \lim_{x\rightarrow 0} \dfrac{\frac{f(x)-f(0)}{x-0}}{\frac{g(x)-g(0)}{x-0}}=\dfrac{\displaystyle\lim_{x\rightarrow 0} \frac{f(x)-f(0)}{x-0}}{\displaystyle\lim_{x\rightarrow 0} \frac{g(x)-g(0)}{x-0}}=\frac{f'(0)}{g'(0)},$$ where you can take the limit 'inside' the fraction because $$f'(0)$$ and $$g'(0)$$ exist and $$g'(0)\neq 0$$. Therefore, it must happen that $$\lim_{x\rightarrow 0} \frac{f'(x)}{g'(x)} = \frac{f'(0)}{g'(0)}.$$
By the Mean Value Theorem, one can prove that $$g(x)\neq 0$$ for $$x\neq 0$$. By the same theorem, for each $$x$$ near $$0$$, we can find a real number $$y$$ between $$0$$ and $$x$$ such that $$\frac{f(x)}{g(x)} = \frac{f(x)-f(0)}{g(x)-g(0)}=\frac{f'(y)}{g'(y)}.$$ Since $$\displaystyle\lim_{y\rightarrow 0} \frac{f'(y)}{g'(y)}$$ exists, we can conclude that $$\lim_{x\rightarrow 0} \frac{f(x)}{g(x)} = \lim_{x\rightarrow 0} \frac{f'(x)}{g'(x)}.$$