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Math Help - L'Hospital's Rule

  1. #1
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    L'Hospital's Rule

    Ok new chapter. I think I under stand this stuff to a point. I have :

    lim x -> 0 e^x -1 -x -(x^2/2)/ x^3 so eventually after taking the derivative of the top and bottom about 3 times i ended up with e^x/6 so the limit approaches infinite correct?

    Then

    lim x -> 0 (1+a/x)^bx, this one is slightly harder i can get the first derivative then i'm not sure what to do. So for the first derivative I have b(1+a/x)^bx-1*(a/x^2) then i'm lost. {edit} ok i just realized different form g(x) ln f(x) so if i do that i get b * ( -a/x^2) so -ab/x^2 would that be negative infinite then?
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  2. #2
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    Quote Originally Posted by UMStudent View Post
    Ok new chapter. I think I under stand this stuff to a point. I have :

    lim x -> 0 e^x -1 -x -(x^2/2)/ x^3 so eventually after taking the derivative of the top and bottom about 3 times i ended up with e^x/6 so the limit approaches infinite correct?
    Look right.

    Then
    lim x -> 0 (1+a/x)^bx, this one is slightly harder i can get the first derivative then i'm not sure what to do. So for the first derivative I have b(1+a/x)^bx-1*(a/x^2) then i'm lost. {edit} ok i just realized different form g(x) ln f(x) so if i do that i get b * ( -a/x^2) so -ab/x^2 would that be negative infinite then?
    If a,b>0

    (1+ax)^(bx)=exp(ln(1+ax)^(bx)) for x>0.
    But then,
    exp(bx*ln(1+ax))

    Note that,

    bx*ln(1+ax) = ln(1+ax)/(1/bx)
    You can use L-Hopital Rule on this.
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  3. #3
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    so then ln(1+ax)/(1/bx) the derivative using L'Hospital's rule is (-a/x^2)/(1+ax)/(b), then from there how do i solve the limit as x -> infinite?
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