0
$\begingroup$

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?

$\endgroup$
2
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Ah! I see! Thank you! $\endgroup$
    – MasonTep
    Mar 7 at 23:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.