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Math Help - lim x->infinity (1+(1/x))^x

  1. #1
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    lim x->infinity (1+(1/x))^x

    Hi,

    I have another question on limits:

    lim (1+(1/x))^x
    x->infinity

    So i assume that I have to use L'Hopital's Rule to get to an answer

    Firstly I have let y=the equation and taken the natural log of each side:

    ln(y)=ln[(1/(1+(1/x)))^x]

    and then made it the form:

    =lim x*ln(1/(1+(1/x)))
    x->infinity

    =lim ln(1+(1+(1/x)))/x^-1
    x->infinity

    and used L'hopital to get to:

    1/(1+(1/x))

    Everywhere i have looked has said that the answer is that x approaches e and I have absolutely no idea how I get this solution? If you are able to give me a hand it would be greatly appreciated! Thanks, sorry it is so messy I don't know how to write it out in the way I have seen some of the answers written.
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  2. #2
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    Re: lim x->infinity (1+(1/x))^x

    By definition :

    e=\lim_{x \to \infty}\left(1+\frac{1}{x}\right)^x
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  3. #3
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    Re: lim x->infinity (1+(1/x))^x

    Quote Originally Posted by princeps View Post
    By definition :

    e=\lim_{x \to \infty}\left(1+\frac{1}{x}\right)^x
    Thanks, although i think the whole question is about proving that that's true through using L' Hopital's Rule
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  4. #4
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    Re: lim x->infinity (1+(1/x))^x

    Quote Originally Posted by Cotty View Post
    Hi,

    I have another question on limits:

    lim (1+(1/x))^x
    x->infinity

    So i assume that I have to use L'Hopital's Rule to get to an answer

    Firstly I have let y=the equation and taken the natural log of each side:

    ln(y)=ln[(1/(1+(1/x)))^x]

    and then made it the form:

    =lim x*ln(1/(1+(1/x)))
    x->infinity

    =lim ln(1+(1+(1/x)))/x^-1
    x->infinity

    and used L'hopital to get to:

    1/(1+(1/x))

    Everywhere i have looked has said that the answer is that x approaches e and I have absolutely no idea how I get this solution? If you are able to give me a hand it would be greatly appreciated! Thanks, sorry it is so messy I don't know how to write it out in the way I have seen some of the answers written.
    \displaystyle \begin{align*} \lim_{x \to \infty} \left[ \left( 1 + \frac{1}{x} \right)^x \right] &= \lim_{x \to \infty} e^{\ln{\left[ \left( 1 + \frac{1}{x} \right)^x \right]}} \\ &= \lim_{x \to \infty} e^{\frac{\ln{\left( 1 + \frac{1}{x} \right)}}{\frac{1}{x}}} \\ &= e^{\lim_{x \to \infty} \frac{\ln{\left( 1 + \frac{1}{x} \right)} }{\frac{1}{x}}} \end{align*}

    This limit goes to \displaystyle \frac{0}{0} so now use L'Hospital's Rule.
    Thanks from ILikeSerena, Cotty and Educated
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  5. #5
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    Re: lim x->infinity (1+(1/x))^x

    Quote Originally Posted by Prove It View Post
    \displaystyle \begin{align*} \lim_{x \to \infty} \left[ \left( 1 + \frac{1}{x} \right)^x \right] &= \lim_{x \to \infty} e^{\ln{\left[ \left( 1 + \frac{1}{x} \right)^x \right]}} \\ &= \lim_{x \to \infty} e^{\frac{\ln{\left( 1 + \frac{1}{x} \right)}}{\frac{1}{x}}} \\ &= e^{\lim_{x \to \infty} \frac{\ln{\left( 1 + \frac{1}{x} \right)} }{\frac{1}{x}}} \end{align*}

    This limit goes to \displaystyle \frac{0}{0} so now use L'Hospital's Rule.

    So is it correct that I used L'Hops to get x/(x+1) and again to get 1/1 and then put e^(1/1)?
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  6. #6
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    Re: lim x->infinity (1+(1/x))^x

    Quote Originally Posted by Cotty View Post
    So is it correct that I used L'Hops to get x/(x+1) and again to get 1/1 and then put e^(1/1)?
    Yes that's correct.
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  7. #7
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    Re: lim x->infinity (1+(1/x))^x

    Thanks Prove It, you've been really helped me today!
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