anonymous
  • anonymous
how do you evaluate the lim h->0 [(1+h)^ln(l+h) -1]/ h
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
L'hospital's rule: indeterminate form of 0/0 Does this help?
dumbcow
  • dumbcow
i agree, differentiate top and bottom and i think limit is zero
anonymous
  • anonymous
So derive the top and bottom seperately

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anonymous
  • anonymous
Does the (1+h)^(ln(1=h) go to 0 already or do I have to rearrange ?
anonymous
  • anonymous
before I can take the derivative
anonymous
  • anonymous
or does it go to one making it 1-1 so i have a 0/0 function
anonymous
  • anonymous
the minus 1 makes it go to 0
anonymous
  • anonymous
so (1+h)^(ln(1=h)= 1 but there is a minus 1 in the numerater so its 0
anonymous
  • anonymous
Medal pleaser ^.^
anonymous
  • anonymous
Ohh ok thanks
anonymous
  • anonymous
so you should show your work, and u'll get 0
anonymous
  • anonymous
Im slightly confused on how to take the derivative of (1+h)^ln(1+h)
dumbcow
  • dumbcow
\[y = (1+h)^{\ln 1+h} \rightarrow \ln y=\ln (1+h)^{\ln 1+h}\] 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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