L'Hopital rule variation

Question

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.

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. — Jason, Oct 07 '16 at 20:50
Original L'Hopital's rule:
If $\lim_{x \rightarrow a} f(x)=0 \quad$ and $\lim_{x \rightarrow a} g(x)=0 \quad$ and $\lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}=L \quad}$ then $\lim_{x \rightarrow a} \frac{f(x)}{g(x)}=L \quad$ and it has a reasoning. — shapoor, Oct 07 '16 at 20:57
Any formulation of L'Hopital's rule I've ever seen says that $lim_{x\to a}f(x)=\lim_{x\to a}g(x)$ and both are equal to either $0$ or $\pm\infty$. — Jason, Oct 07 '16 at 20:59
If $\quad \lim_{x \rightarrow a} f(x)=0 \quad $ and $\quad \lim_{x \rightarrow a} g(x)=0 \quad$ and $\quad \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}=L, \quad $ then $$\lim_{x \rightarrow a} \frac{f(x)}{g(x)}=L $$ — shapoor, Oct 07 '16 at 21:06
@shapoor This is as well one of the cases in which L'Hopital's can be applied. — Kat, Oct 07 '16 at 21:11
the proof for the question. because I know the proof of the original statement, which I mentioned in the comments. — shapoor, Oct 07 '16 at 21:16
@shapoor The are already answers you can look at Case infinity over infinity. Case 0/0 — Kat, Oct 07 '16 at 21:28

Kat · Answer 1 · 2016-10-07T20:48:42.693

1

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.

edited Oct 07 '16 at 20:48

answered Oct 07 '16 at 20:41

Kat

706

Reasoning please. – shapoor Oct 07 '16 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 07 '16 at 20:46

score 0 · Answer 2 · answered Oct 07 '16 at 20:28

0

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

answered Oct 07 '16 at 20:28

user333036

31

2

The first does not imply the second. – H. H. Rugh Oct 07 '16 at 20:37

score 0 · Answer 3 · answered Oct 08 '16 at 07:29

0

Which is exactly right and what I explicitly stated earlier

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

answered Oct 08 '16 at 07:29

user333036

31

L'Hopital rule variation

3 Answers3