Question

Prove l'hopital's rule for the indeterminant form 0/0 as x--> infinity+

Answer 1

Thm: Suppose \[\lim_{x \rightarrow \infty} f(x) =\lim_{x \rightarrow \infty} g(x)=0,\] that f' and g' exist and are continuous on some open interval \[x >\frac{ a }{ 2 }>0\], that g'(x)\[\neq0\] for \[x >\frac{ a }{ 2 }\] and that \[\lim_{x \rightarrow \infty} \frac{ f'(x) }{ g'(x) }=L\] exists and is finite. Then, \[\lim_{x \rightarrow \infty}\frac{ f(x) }{ g(x) }=\lim_{x \rightarrow \infty}\frac{ f'(x) }{ g'(x) }\]

Answer 2

does this help
http://mathforum.org/library/drmath/view/53340.html

Answer 3

sort of, though from what I can see that proves the indeterminant form 0/0 as x approaches a finite value, not infinity

Answer 4

look closer, it does both 0/0 and inf/inf

Answer 5

I see that it proves for 0/0 and inf/inf. But it takes the lim x-->b, not x-->inf. Unless you're seeing something that I dont

Answer 6

to prove it for x-->inf you can make a simple modification to that proof:
lim x-->inf (f(x)) = lim y-->0+ (f(1/y))

Answer 7

ok, I jumped to the conclusion you wanted inf/inf.... you are doing a different problem

Answer 8

@cnknd Ok, I see that as the last example on the link. Is that proof good for the 0/0 form? I'm confused bc I feel like I have to use the a/2 somewhere (It is given in the thm), but I am not sure what to do with it. If i need to do anything with it.

Answer 9

sry, i have no clue what that a/2 in the thm means

Answer 10

Ok, I will just keep trying to work it out. Thanks