I have a question involving limits

Question

anonymous · Answer

yeah

anonymous · Answer

I'm sure you do. . . ! Lol.

anonymous · Answer

$$ \lim_{h \rightarrow 0 } (1+h)^\ln(1+h)-1/h$$

yuki · Answer

shoot it

anonymous · Answer

the ln(1=h) is all together as a power

anonymous · Answer

the ln(1+h) rather

anonymous · Answer

it looks like the definition of a derivative... almost

yuki · Answer

is it $$\lim_{h \rightarrow 0} ((1+h)^{\ln(1+h)}-1 )/h$$

?

anonymous · Answer

someone asked this earlier. but use L'hopital's rule and get an indeterminate form of 0/0 by deriving the numerater and denomenator seperately

anonymous · Answer

yes

anonymous · Answer

do that and u'll get ur answer. Tell me what you get cuz I don't wanna just give u the answer

anonymous · Answer

sounds good to me!

anonymous · Answer

medal pwease

yuki · Answer

lol

anonymous · Answer

i got it to be 0/0 but i don't know where to go exactly from there

anonymous · Answer

thats ur answer

anonymous · Answer

ohhh, that easy haha. thanks pointing me in the right direction!

anonymous · Answer

0/0 is the answer?

anonymous · Answer

No, 0/0 is an indeterminate answer. If you get a limit in the form of 0/0, you have to use l'Hopital's rule -- if  limit of f(x)/g(x) as x ----> infinity = 0/0 or infinity/infinity, then that same limit equals the limit of f'(x)/g'(x). (The limit of the ratio of two functions giving 0/0 or infinity/infinity = the limit of the ratio of their derivatives).

anonymous · Answer

looks like its back to the drawing board for me then!