suppose L'Hopital applies and $$\lim_{x\to\infty}\frac{f(x)}{g(x)} = \lim_{x\to\infty}\frac{f'(x)}{g'(x)}$$ under what conditions is it true then that

$$\lim_{x\to\infty}\frac{\frac{f(x)}{g(x)} }{ \frac{f'(x)}{g'(x)}}=k$$

for non-zero constant $k$

background: consider the sum $$\sum\frac{f(n)}{g(n)}$$ for example, $$\sum \frac{n}{n^3+5}$$ and consider a student's approach taking the limit on the nth term using l'hopital $$\lim_{x\to\infty} \frac{n}{n^3+5}=\lim_{x\to\infty}\frac{1}{3n^2}$$

at this point the student declares the series convergent as it behaves like $\sum \frac{1}{3n^2}$. Obviously the LCT is not being used in the traditional sense, yet there may be something true in this madness.. I suspect..

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    $\begingroup$ Well, in the case of $\lim_{x\to\infty}\frac{n}{n^3}$ this obviously doesn't work, since $\frac{\frac{f(x)}{g(x)} }{ \frac{f'(x)}{g'(x)}} = \frac{1/n^2}{1/3n^2}= 3\neq 1$. $\endgroup$ – Arthur Apr 20 '14 at 7:25
  • $\begingroup$ thank you @Arthur, i corrected that... i am specially interested in when the limit comparison test would apply using the post l'hopital function as a series to compare to..thanks for your comment, I changed the 1 to a $k$ $\endgroup$ – userX Apr 20 '14 at 7:30
  • $\begingroup$ Isn't it always going to be equal to a constant? If the original ratio diverges then the ratio of derivatives will also diverge (=> constant), and so will it be is the original ratio converges. $\endgroup$ – Aegis Apr 20 '14 at 19:13
  • $\begingroup$ incidentally; i found the answer here math.stackexchange.com/questions/39979/… $\endgroup$ – userX Apr 21 '14 at 0:15
  • $\begingroup$ and also here math.stackexchange.com/questions/77024/… $\endgroup$ – userX Apr 21 '14 at 0:16

Is 0 = 0 (Yes True)

Is 0/0 = 1 (I don’t think so)(You can not divide by zero)

and for convergence the series Σ(n/(n^3 +5)) is convergent if Σ(1/n^2) is as

n/(n^3 +5) is less than (n/n^3)

An advice: Whenever using any rule don’t be afraid of it , just try and understand what it says.Now L’Hospitals Rule is based upon what?(xRead Taylor’s Series or google the proof) then you can understand what it means or how it came into being ,what were the assumptions and certainly don’t be afraid if you don’t understand any of the stuff , just be patient maybe give it a day’s thought and come back again to it.


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