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Math Help - help with limit

  1. #1
    Super Member 11rdc11's Avatar
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    help with limit

    I'm trying to figure out how the \lim_{x\to 0}(1+x)^\frac{1}{x}= e. I get the answer to be 1.

    (\frac{1}{x})(\ln{1+x})= \infty\cdot 0 = 0

    so e^0 = 1

    What am I doing wrong? Thanks
    Last edited by 11rdc11; August 25th 2008 at 10:46 PM. Reason: trying to learn latex
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  2. #2
    Super Member wingless's Avatar
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    \infty\cdot 0 is an indeterminate form.

    See Indeterminate form - Wikipedia, the free encyclopedia.
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  3. #3
    Super Member 11rdc11's Avatar
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    Thanks and i just covered that section lol. I just made the assumption that 0 multiplied by anything is 0.
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  4. #4
    Eater of Worlds
    galactus's Avatar
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    One way is to build on the differentiability of ln(x).

    More specifically,on the derivative of ln(x) at x=1.

    \frac{d}{dx}[ln(x)]\left|=\frac{1}{x}\right|_{x=1}=1

    If we express this using the definition of a derivative, we get:

    1=\lim_{h\to 0}\frac{ln(1+h)-ln(1)}{h}=\lim_{h\to 0}\frac{ln(1+h)}{h}=\lim_{h\to 0}ln(1+h)^{\frac{1}{h}}

    And it follows that we get:

    \displaystyle{e=e^{\left(\displaystyle{\lim_{h\to 0}ln(1+h)^{\frac{1}{h}}}\right)}}

    Now, from the continuity of e, we can write:

    \displaystyle{e=\lim_{h\to 0}e^{ln(1+h)^{\frac{1}{h}}}=\lim_{h\to 0}(1+h)^{\frac{1}{h}}}
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