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  1. #1
    Member akhayoon's Avatar
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    limit

    \lim{x}\rightarrow{\infty} \frac{(1+x)^{\frac{1}{x}}-e}{x}

    well the answer is -e/2, but I can't seem to get the answer...mostly due the fact that there is a 1/x power on the 1+x polynomial and how to implement it with
    L'Hopitale's rule...?
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  2. #2
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    Quote Originally Posted by akhayoon View Post
    \lim{x}\rightarrow{\infty} \frac{(1+x)^{\frac{1}{x}}-e}{x}

    well the answer is -e/2, but I can't seem to get the answer...mostly due the fact that there is a 1/x power on the 1+x polynomial and how to implement it with
    L'Hopitale's rule...?
    Well, \lim_{x \rightarrow \infty} (1 + x)^{1/x} = e so you can use l'Hopital's rule.

    To get the derivative of y = (1 + x)^{1/x} you can use logarithmic differentiation.

    Edit: My mistake: \lim_{x \rightarrow \infty} (1 + x)^{1/x} = 1, NOT e. l'H CANNOT be used. See post #4.
    Last edited by mr fantastic; March 8th 2008 at 05:10 PM.
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  3. #3
    Member akhayoon's Avatar
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    still kind of shaky on this I tried multiplying everything by ln so I could reduce the
    (1+x)^{\frac{1}{x}} with the ln rule, but everything gets kind of messy and I end up getting a function that could never give me e
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  4. #4
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    Quote Originally Posted by akhayoon View Post
    still kind of shaky on this I tried multiplying everything by ln so I could reduce the
    (1+x)^{\frac{1}{x}} with the ln rule, but everything gets kind of messy and I end up getting a function that could never give me e
    I made a careless mistake in the limit and have edited my reply. As posted, the value of the limit is zero. Have you posted the correct expression? Perhaps x --> 0, not oo?
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  5. #5
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    Quote Originally Posted by akhayoon View Post
    still kind of shaky on this I tried multiplying everything by ln so I could reduce the
    (1+x)^{\frac{1}{x}} with the ln rule, but everything gets kind of messy and I end up getting a function that could never give me e
    The limit >is< meant to be x --> 0. \lim_{x \rightarrow 0} (1+x)^{\frac{1}{x}} = e so l'Hospital's Rule can be used.

    To get the derivative of y = (1 + x)^{1/x}:

    \ln y = \frac{1}{x} \ln (1 + x)


    \Rightarrow \frac{1}{y}\, \frac{dy}{dx} = - \frac{1}{x^2} \ln (1 + x) + \frac{1}{x(1 + x)}


    \Rightarrow \frac{dy}{dx} = y \left( - \frac{1}{x^2} \ln (1 + x) + \frac{1}{x(1 + x)} \right).


    So the limit becomes:


    \lim_{x \rightarrow 0} (1 + x)^{1/x} \left( - \frac{1}{x^2} \ln (1 + x) + \frac{1}{x(1 + x)} \right) = e \lim_{x \rightarrow 0} \left( - \frac{1}{x^2} \ln (1 + x) + \frac{1}{x(1 + x)} \right).


    Now show that \lim_{x \rightarrow 0} \left( - \frac{1}{x^2} \ln (1 + x) + \frac{1}{x(1 + x)} \right) = \lim_{x \rightarrow 0} \left(\frac{x - (1 + x) \, \ln (1 + x)}{x^2(1 + x)} \right) = - \frac{1}{2}.

    Multiple applications of l'Hospital's rule is one possible approach.
    Another approach is to substitute the series expansion for ln (1 + x): \ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - ....... This substitution is valid since the series converges for -1 < x \leq 1 and x \rightarrow 0 ......
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  6. #6
    Member akhayoon's Avatar
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    ok, so I evaluate the limit as it approaches zero to get -e/2 like u showed, but aren't I asked to evaluate it as it approaches infinity? so how does this cover that?
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  7. #7
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    Quote Originally Posted by akhayoon View Post
    ok, so I evaluate the limit as it approaches zero to get -e/2 like u showed, but aren't I asked to evaluate it as it approaches infinity? so how does this cover that?
    I have already addressed that question - read post #4 again: In particular the bit where I say that if x --> oo the limit is equal to 0.

    Clearly the limit is meant to be x --> 0. Clearly there is a typographical error in the question.
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