# Continuous functions quotient derivative at $0$

This problem was on an exam I took (I have tried to remember it how it was but I don't have the original transcript).

Let $$X$$ be a metric space and let $$f_1,f_2:X\rightarrow \mathbb{R}$$ be two continuous differentiable functions. Suppose at some point $$x_0$$, $$f_1(x_0)=0=f_2(x_0)$$ and $$f'_1(x_0)=C_1$$ and $$f'_2(x_0)=C_2$$ for some non-zero finite constants $$C_1$$ and $$C_2$$. Prove that:

$$\lim_{x\rightarrow x_0}\frac{f_1(x)}{f_2(x)}=K$$

for some finite constant $$K$$.

The exam is over and I got my mark but they don't give us the answers and it has bugged me since. Noone I know seems to have been able to answer it. Could anyone help me with this?

• Mar 7 at 22:25
• @MartínVacasVignolo "$f'_2(x_0)=C_2$ for some non-zero finite constants $C_1$ and $C_2$" implies $f_2$ is not identically $0$. Mar 7 at 22:47

Directly applying the limit $$x \rightarrow x_0$$ yields the indeterminate form $$\frac{0}{0}$$. Thus (and since your functions are continuous and differentiable) you may then apply L'Hopital's Rule to obtain a quotient of the derivatives. Applying the limit now yields a constant (no longer an indeterminate form):
$$\lim_{x\rightarrow x_0}\frac{f_1(x)}{f_2(x)}=\lim_{x\rightarrow x_0}\frac{f'_1(x)}{f'_2(x)}=\frac{f'_1(x_0)}{f'_2(x_0)}=\frac{C_1}{C_2} \equiv K$$