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anonymous
 5 years ago
how do you evaluate the lim h>0 [(1+h)^ln(l+h) 1]/ h
anonymous
 5 years ago
how do you evaluate the lim h>0 [(1+h)^ln(l+h) 1]/ h

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0L'hospital's rule: indeterminate form of 0/0 Does this help?

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.1i agree, differentiate top and bottom and i think limit is zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So derive the top and bottom seperately

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Does the (1+h)^(ln(1=h) go to 0 already or do I have to rearrange ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0before I can take the derivative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or does it go to one making it 11 so i have a 0/0 function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the minus 1 makes it go to 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so (1+h)^(ln(1=h)= 1 but there is a minus 1 in the numerater so its 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you should show your work, and u'll get 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Im slightly confused on how to take the derivative of (1+h)^ln(1+h)

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.1\[y = (1+h)^{\ln 1+h} \rightarrow \ln y=\ln (1+h)^{\ln 1+h}\] use log rule log x^n = n*log x \[\ln y=(\ln (1+h))^{2}\] now differentiate \[\frac{1}{y}\frac{dy}{dx} = \frac{2}{1+h} \ln (1+h)\] multiply by y on both sides where y=(1+h)^ln(1+h) \[\frac{dy}{dx} = 2\ln (1+h)*(1+h)^{\ln (1+h) 1}\]
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