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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm sure you do. . . ! Lol.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[ \lim_{h \rightarrow 0 } (1+h)^\ln(1+h)1/h\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the ln(1=h) is all together as a power

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it looks like the definition of a derivative... almost

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is it \[\lim_{h \rightarrow 0} ((1+h)^{\ln(1+h)}1 )/h\] ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0someone 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
 5 years ago
Best ResponseYou've already chosen the best response.0do that and u'll get ur answer. Tell me what you get cuz I don't wanna just give u the answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got it to be 0/0 but i don't know where to go exactly from there

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohhh, that easy haha. thanks pointing me in the right direction!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, 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
 5 years ago
Best ResponseYou've already chosen the best response.0looks like its back to the drawing board for me then!
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