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Math Help - Show definition of derivative

  1. #1
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    Show definition of derivative

    Show that lim_{x-->0}  ln (1+x)/x  = 1 from the definition of derivative.
    Last edited by wopashui; January 28th 2010 at 09:16 PM.
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  2. #2
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    Quote Originally Posted by wopashui View Post
    Show that lim_x->0  ln (1+x)/x = 1 from the definition of derivative.
    Hello
    \lim_{x\to 0} \frac{ln(1+x)}{x} = \lim_{x\to 0} \frac{ln(1+x)-0}{x-0}
    Let f(x)=ln(1+x)
    Clearly f(0)=0
    So our limit can be written as:
    \lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=f'(0)

    Got it?
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  3. #3
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    Quote Originally Posted by General View Post
    Hello
    \lim_{x\to 0} \frac{ln(1+x)}{x} = \lim_{x\to 0} \frac{ln(1+x)-0}{x-0}
    Let f(x)=ln(1+x)
    Clearly f(0)=0
    So our limit can be written as:
    \lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=f'(0)

    Got it?

    so that produce f(x)/x, why is it equal to 1?
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  4. #4
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    Quote Originally Posted by wopashui View Post
    so that produce f(x)/x, why is it equal to 1?
    I did not follow you!
    \frac{ln(1+x)}{x}=\frac{ln(1+x)-0}{x-0}
    Do you have problem in this one?
    Substracting zero is allowed everywhere.
    and I replaced 0 by f(0)
    since f(0)=ln(1+0)=ln(1)=0

    Please, explain more.
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  5. #5
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    Quote Originally Posted by General View Post
    I did not follow you!
    \frac{ln(1+x)}{x}=\frac{ln(1+x)-0}{x-0}
    Do you have problem in this one?
    Substracting zero is allowed everywhere.
    and I replaced 0 by f(0)
    since f(0)=ln(1+0)=ln(1)=0

    Please, explain more.
    I meant, we are proving the limit is equal to 1 not 0, how does that make sense?
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