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Math Help - Limits

  1. #1
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    Limits

    Please help me to find out the following limits.
    1. lim {(1+1/x)^x} as x->0

    2. lim{(a^x - 1)/x} as x->(+infinity), a>1

    Dear friends,
    Please try them out without using L'Hopital's rule.
    Last edited by nirmalya basu; August 9th 2006 at 07:09 AM.
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  2. #2
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    Hello, nirmalya basu!

    I can help with #1 if you are allowed to use L'Hopital . . .


    1)\;\;\lim_{x\to0}\left(1 + \frac{1}{x}\right)^x

    Let y\;=\;\left(1 + \frac{1}{x}\right)^x

    Take logs: . \ln y\;=\;\ln\left(1 + \frac{1}{x}\right)^x \;= \;x\cdot\ln\left(1 + \frac{1}{x}\right)

    Let x = \frac{1}{u} . . . Note that, if x\to0, then u \to \infty

    We have: . \ln y \;= \;\frac{1}{u}\ln(1 + u)\;= \;\frac{\ln(1 + u)}{u}

    If u\to\infty, we have the indeterminate form: \frac{\infty}{\infty}
    . . Hence, we may use L'Hopital.

    Then we have: . \ln y\;=\;\frac{\frac{1}{1 + u}}{1}\;=\;\frac{1}{1 + u}

    Hence: . \lim_{u\to\infty}(\ln y) \;= \;\lim_{u\to\infty}\left(\frac{1}{1 + u}\right)\;=\;0


    Since \lim_{u\to\infty}(\ln y)\:=\:0, then \lim_{u\to\infty} y \:=\:e^0 \:=\:1


    Therefore: . \lim_{x\to\0}\left(1 + \frac{1}{x}\right)^x\;=\;1

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  3. #3
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    Quote Originally Posted by Soroban
    Hello, nirmalya basu!

    I can help with #1 if you are allowed to use L'Hopital . . .
    There is still another detail missing. You assumed the limit exists! (You probably realized that, right? You just did not mention it cuz to make your work easier to follow).
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