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Math Help - Limit : ln-expo

  1. #1
    Super Member dhiab's Avatar
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    Limit : ln-expo

    Calculate :
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  2. #2
    Moo
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    Quote Originally Posted by dhiab View Post
    Calculate :
    \lim_{x\to 0} \frac{e^x-e^0}{\ln(x+1)-\ln(0+1)}=\lim_{x\to 0}\frac{e^x-e^0}{x}\cdot\frac{x}{\ln(1+x)-\ln(1+0)}=\frac{f'(0)}{g'(0)}

    where f(x)=e^x and g(x)=ln(1+x)
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  3. #3
    Math Engineering Student
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    the limit is 1 because \underset{x\to 0}{\mathop{\lim }}\,\frac{e^{x}-1}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\ln (1+x)}{x}=1.
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  4. #4
    Senior Member nikhil's Avatar
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    Lightbulb aliter

    same can be done using L'Hospital rule (if on putting limit you get a indeterminate form like 0/0 or infinity/infinity, then keep on differentiating both numerator and denominator until you get a fininte limit)

    A=lim (e^x-1)/[ln(x+1)]
    x->0
    on putting x=0 we get
    A=(1-1)/ln(1) =0/0
    so differentiating both numerator and denominator we get
    A=lim (e^x)(x+1)
    x->0
    now putting x=0
    we get A=(e^0)1=1
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  5. #5
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    Quote Originally Posted by dhiab View Post
    Calculate :
    Since this question is posted in the PRE-Calculus subforum, I assume a solution that doesn't use calculus is required ....?


    Edit: All subsequent off-topic posts in this thread have been moved to here: http://www.mathhelpforum.com/math-he...continues.html. Please feel free to continue the discussion in the Chatroom.
    Last edited by mr fantastic; June 23rd 2009 at 02:35 AM.
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