# L'Hopital rule variation [duplicate]

If $$\lim_{x \rightarrow a} f(x)= \infty\quad \lim_{x \rightarrow a} g(x)=\infty$$ and $$\lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}=L$$ then
$$\lim_{x \rightarrow a} \frac{f(x)}{g(x)}=L$$

Is this correct? Any response would be appreciated.

• This is just L'Hopital's rule? Oct 7, 2016 at 20:47
• @Jason Indeed it is.
– Kat
Oct 7, 2016 at 20:48
• But I mean, the title of the question is "L'Hospital's role [sic] variation", but this is not a variation, this is just literally L'Hopital's rule. Oct 7, 2016 at 20:50
• @Jason you're right
– Kat
Oct 7, 2016 at 20:51
• @shapoor The are already answers you can look at Case infinity over infinity. Case 0/0
– Kat
Oct 7, 2016 at 21:28

Note that $\lim_{x\rightarrow a} \frac {f(x)}{g(x)}$ give you an indeterminate form "$\frac {\infty}{\infty}$" and from here L'Hopital rule can be applied, and since you know: $\lim_{x\rightarrow a} \frac {f'(x)}{g'(x)}=L \Rightarrow \lim_{x\rightarrow a} \frac {f(x)}{g(x)}=L$ is correct assuming L is a finite value.

• Reasoning please. Oct 7, 2016 at 20:42
• I'm just using the fact fact that $lim_{x\rightarrow a} f(x) = \infty$ and $lim_{x\rightarrow a} g(x) = \infty$ therefore $lim_{x\rightarrow a} \frac {f(x)}{g(x)}$ is an indeterminate form "$\frac {\infty}{\infty}$" which is one of the cases in which you can apply L'Hopital's rule
– Kat
Oct 7, 2016 at 20:46

Yes. This is correct according to the theorem.

If lim x-> a F (x)/g(x) = L

Hence lim x-> a f '(x)/g '(x) =L

And vice versa

• The first does not imply the second. Oct 7, 2016 at 20:37

Which is exactly right and what I explicitly stated earlier

0/0 and inf/inf are indeterminate forms hence Lhopital stands as quoted