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Math Help - Evaluation of limit (2)

  1. #1
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    Evaluation of limit (2)

    pliz help me out with this,

    lim_{x\rightarrow 0} \frac{(1+x)^{1/x}-e}{x}
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  2. #2
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    Re: Evaluation of limit (2)

    what happens if you make the substitution n = 1/x?
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    Re: Evaluation of limit (2)

    ok..after replacing n=1/x....i gt
    lim_{n\rightarrow inf} \frac{(1+1/n)^{n}-e} {n}...
    what to do next...
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    Re: Evaluation of limit (2)

    Quote Originally Posted by rishilaish View Post
    ok..after replacing n=1/x....i gt
    lim_{n\rightarrow inf} \frac{(1+1/n)^{n}-e} {n}...
    what to do next...
    \lim_{n \to \infty} \left(1+\dfrac{1}{n}\right)^n = e

    This is a common limit well worth knowing.


    Edit: you need to change your denominator too: n = \dfrac{1}{x} \Leftrightarrow x = \dfrac{1}{n}

    \lim_{n \to \infty} \left(\dfrac{\left(1+\frac{1}{n}\right)^n - e}{\frac{1}{n}}\right)


     = \lim_{n \to \infty} \left(n \left(1+\frac{1}{n}\right)^n - e\right)
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  5. #5
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    Re: Evaluation of limit (2)

    Quote Originally Posted by rishilaish View Post
    ok..after replacing n=1/x....i gt
    lim_{n\rightarrow inf} \frac{(1+1/n)^{n}-e} {n}...
    what to do next...
    It would actually be \displaystyle \lim_{x \to \infty}\frac{\left(1 + \frac{1}{n}\right)^n - e}{\frac{1}{n}}

    Personally I don't see making the substitution \displaystyle n = \frac{1}{x} as necessary, since the function already went to \displaystyle \frac{0}{0} and the denominator of the original function is easier to differentiate so that you can use L'Hospital's Rule...
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  6. #6
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    Re: Evaluation of limit (2)

    Quote Originally Posted by e^(i*pi) View Post
    \lim_{n \to \infty} \left(1+\dfrac{1}{n}\right)^n = e

    This is a common limit well worth knowing.


    Edit: you need to change your denominator too: n = \dfrac{1}{x} \Leftrightarrow x = \dfrac{1}{n}

    \lim_{n \to \infty} \left(\dfrac{\left(1+\frac{1}{n}\right)^n - e}{\frac{1}{n}}\right)


     = \lim_{n \to \infty} \left(n \left(1+\frac{1}{n}\right)^n - e\right)
    i think

    \lim_{n \to 0} \left(1+\dfrac{1}{n}\right)^n = e
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  7. #7
    MHF Contributor Siron's Avatar
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    Re: Evaluation of limit (2)

    Quote Originally Posted by rishilaish View Post
    i think

    \lim_{n \to 0} \left(1+\dfrac{1}{n}\right)^n = e

    No, n has to approach to \infty. But this limit is not easy, because if you follow the suggestion of Prove It by applying l'Hopital's rule then the derivative of (1+x)^{\frac{1}{x}} is not very easy and I don't know if it simplifies the problem.
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  8. #8
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    Re: Evaluation of limit (2)

    Quote Originally Posted by rishilaish View Post
    i think

    \lim_{n \to 0} \left(1+\dfrac{1}{n}\right)^n = e

    That's not the case, it is definitely the limit to infinity: lim n to infinity (1+1/n)^n - Wolfram|Alpha

    If you want choose a suitable large value of n and compare it to e.
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  9. #9
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    Re: Evaluation of limit (2)

    somebody plz solve it...the answer is -e/2
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  10. #10
    MHF Contributor Siron's Avatar
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    Re: Evaluation of limit (2)

    By first inspection you notice you get the indeterminate form \frac{0}{0} therefore you can apply l'Hopital's rule, that means:
    \lim_{x \to 0} \frac{\frac{d}{dx}(1+x)^{\frac{1}{x}}-\frac{d(e)}{dx}}{\frac{d(x)}{dx}}
    =\lim_{x\to 0} \frac{d}{dx}(1+x)^{\frac{1}{x}}

    Calculating \frac{d}{dx}(1+x)^{\frac{1}{x}} gives:
    \frac{d}{dx}e^{\frac{\ln(1+x)}{x}}
    =e^{\frac{\ln(1+x)}{x}}\cdot \frac{d}{dx}\left[\frac{\ln(1+x)}{x}\right]

    Calculating \frac{d}{dx}\left[\frac{\ln(1+x)}{x}\right] gives:
    \frac{\frac{x}{1+x}-\ln(1+x)}{x^2}

    So now you have this limit:
    \lim_{x\to 0} e^{\frac{\ln(1+x)}{x}}\cdot \frac{\frac{x}{1+x}-\ln(1+x)}{x^2}

    which is equal to (because the limit of a product is the product of the limits of the separated factors):
    \lim_{x\to 0} e^{\frac{\ln(1+x)}{x}} \cdot \lim_{x\to 0} \frac{\frac{x}{1+x}-\ln(1+x)}{x^2}

    And
    \lim_{x\to 0} e^{\frac{\ln(1+x)}{x}}=e (recheck that!)

    Therefore the only thing we have to do is to calculate:
    \lim_{x\to 0} \frac{\frac{x}{1+x}-\ln(1+x)}{x^2}
    You notice you get again the indeterminate form \frac{0}{0} therefore we can apply l'Hopital's rule again:
    \lim_{x\to 0} \frac{\frac{x}{1+x}-\ln(1+x)}{x^2}=\lim_{x\to 0} \frac{\frac{1}{(1+x)^2}-\frac{1}{1+x}}{2x}
    =\lim_{x \to 0} \frac{-x}{(1+x)^2\cdot 2x}=\lim_{x\to 0}\frac{-1}{2(1+x)^2}=\frac{-1}{2}

    Therefore the limit is \frac{-e}{2}
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  11. #11
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    Re: Evaluation of limit (2)

    thanks siron...u help me out every detail...examz getting close so iv to prepare hard...
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