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?


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):

$$\lim_{x\rightarrow x_0}\frac{f_1(x)}{f_2(x)}=\lim_{x\rightarrow x_0}\frac{f'_1(x)}{f'_2(x)}=\frac{f'_1(x_0)}{f'_2(x_0)}=\frac{C_1}{C_2} \equiv K$$

where the first equality is obtained by L'Hopital's Rule.

  • $\begingroup$ Ah! I see! Thank you! $\endgroup$
    – MasonTep
    Mar 7 at 23:04

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