## anonymous 5 years ago I have a question involving limits

1. anonymous

yeah

2. anonymous

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

3. anonymous

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

4. Yuki

shoot it

5. anonymous

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

6. anonymous

the ln(1+h) rather

7. anonymous

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

8. Yuki

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

9. anonymous

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

10. anonymous

yes

11. anonymous

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

12. anonymous

sounds good to me!

13. anonymous

medal pwease

14. Yuki

lol

15. anonymous

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

16. anonymous

17. anonymous

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

18. anonymous

19. anonymous

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).

20. anonymous

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