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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.

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  • $\begingroup$ This is just L'Hopital's rule? $\endgroup$
    – Jason
    Oct 7, 2016 at 20:47
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    $\begingroup$ @Jason Indeed it is. $\endgroup$
    – Kat
    Oct 7, 2016 at 20:48
  • $\begingroup$ 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. $\endgroup$
    – Jason
    Oct 7, 2016 at 20:50
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    $\begingroup$ @Jason you're right $\endgroup$
    – Kat
    Oct 7, 2016 at 20:51
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    $\begingroup$ @shapoor The are already answers you can look at Case infinity over infinity. Case 0/0 $\endgroup$
    – Kat
    Oct 7, 2016 at 21:28

3 Answers 3

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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.

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  • $\begingroup$ Reasoning please. $\endgroup$
    – shapoor
    Oct 7, 2016 at 20:42
  • $\begingroup$ 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 $\endgroup$
    – Kat
    Oct 7, 2016 at 20:46
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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

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    $\begingroup$ The first does not imply the second. $\endgroup$
    – H. H. Rugh
    Oct 7, 2016 at 20:37
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Which is exactly right and what I explicitly stated earlier

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

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