# L'Hopital quicky

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..

• Well, in the case of $\lim_{x\to\infty}\frac{n}{n^3}$ this obviously doesn't work, since $\frac{\frac{f(x)}{g(x)} }{ \frac{f'(x)}{g'(x)}} = \frac{1/n^2}{1/3n^2}= 3\neq 1$. Apr 20, 2014 at 7:25
• 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 $k$ Apr 20, 2014 at 7:30
• 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. Apr 20, 2014 at 19:13
• incidentally; i found the answer here math.stackexchange.com/questions/39979/… Apr 21, 2014 at 0:15
• and also here math.stackexchange.com/questions/77024/… Apr 21, 2014 at 0:16