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Math Help - limit help

  1. #1
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    limit help

    I got stuck on this limit
    lim as x approaches -1 ln(2+x)/(x+1)
    I got the limit down to lim as x approaches -1 ((1/2+x))/(x+1) by using L'Hospitals rule, but got stuck from there. I am having trouble getting an answer of 1. (answer given in book thanks.)
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  2. #2
    Eater of Worlds
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    You're getting there.

    \lim_{x\rightarrow{-1}}\frac{ln(2+x)}{x+1}

    L'Hopital: \lim_{x\rightarrow{-1}}\frac{1}{x+2}

    Now, see how it's 1?.
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  3. #3
    Math Engineering Student
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    D_x\ln(x+2)=\frac1{x+2} & (x+1)'=1

    Is that what you need?
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  4. #4
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    Quote Originally Posted by davecs77 View Post
    I got stuck on this limit
    lim as x approaches -1 ln(2+x)/(x+1)
    We can avoid the use of L'Hopital's rule.

    First,
    \lim_{x\to -1} \frac{\ln (x+2)}{x+1} = \lim_{x\to 0} \frac{\ln (x+1)}{x}

    Notice the derivative of f(x) = \ln (x+1) at zero is f'(0) = 1.

    But, using the definition of derivative we get,
    \lim_{x\to 0} \frac{\ln (x+1) - \ln (0+1)}{x- 0} = \lim_{x\to 0}\frac{\ln (x+1)}{x} =f'(0)= 1
    Last edited by ThePerfectHacker; July 13th 2007 at 07:41 AM.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    We can avoid the use of L'Hopital's rule.

    First,
    \lim_{x\to -1} \frac{\ln (x+2)}{x+1} = \lim_{x\to 0} \frac{\ln (x+1)}{x}

    Notice the derivative of f(x) = \ln (x+1) at zero is f'(0) = \frac{1}{2}.

    But, using the definition of derivative we get,
    0)= \frac{1}{2}" alt="\lim_{x\to 0} \frac{\ln (x+1) - \ln (0+1)}{x- 0} = \lim_{x\to 0}\frac{\ln (x+1)}{x} =f'(0)= \frac{1}{2}" />
    That last line should read:
    1)= 1" alt="\lim_{x\to 0} \frac{\ln (x+1) - \ln (0+1)}{x- 0} = \lim_{x\to 0}\frac{\ln (x+1)}{x} =f'(1)= 1" />
    as f^{\prime}(1) = 1. This gives us back the correct answer for the limit.

    -Dan
    Last edited by topsquark; July 13th 2007 at 07:50 AM.
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  6. #6
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    Thank you I fixed it.
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