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anonymous

  • 5 years ago

how do you evaluate the lim h->0 [(1+h)^ln(l+h) -1]/ h

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  1. anonymous
    • 5 years ago
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    L'hospital's rule: indeterminate form of 0/0 Does this help?

  2. dumbcow
    • 5 years ago
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    i agree, differentiate top and bottom and i think limit is zero

  3. anonymous
    • 5 years ago
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    So derive the top and bottom seperately

  4. anonymous
    • 5 years ago
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    Does the (1+h)^(ln(1=h) go to 0 already or do I have to rearrange ?

  5. anonymous
    • 5 years ago
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    before I can take the derivative

  6. anonymous
    • 5 years ago
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    or does it go to one making it 1-1 so i have a 0/0 function

  7. anonymous
    • 5 years ago
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    the minus 1 makes it go to 0

  8. anonymous
    • 5 years ago
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    so (1+h)^(ln(1=h)= 1 but there is a minus 1 in the numerater so its 0

  9. anonymous
    • 5 years ago
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    Medal pleaser ^.^

  10. anonymous
    • 5 years ago
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    Ohh ok thanks

  11. anonymous
    • 5 years ago
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    so you should show your work, and u'll get 0

  12. anonymous
    • 5 years ago
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    Im slightly confused on how to take the derivative of (1+h)^ln(1+h)

  13. dumbcow
    • 5 years ago
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    \[y = (1+h)^{\ln 1+h} \rightarrow \ln y=\ln (1+h)^{\ln 1+h}\] 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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