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Math Help - limit

  1. #1
    Super Member dhiab's Avatar
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    limit

    Calculat :

    limit-1_1281982099.jpg
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: limit

    Use L=\lim_{x\to a}f(x)^{g(x)}=e^{\lambda} where \lambda=\lim_{x\to a}\log f(x)^{g(x)} .
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  3. #3
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    Re: limit

    Quote Originally Posted by dhiab View Post
    Calculat :

    Click image for larger version. 

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    Dear dhiab,

    \lim_{x\rightarrow 0}\left(\frac{2-\cos x}{1-\cos x}\right)^{\sin^{2}x}

    =\lim_{x\rightarrow 0}e^{\left(\sin^{2}x\right)\ln\left(\frac{2-\cos x}{1-\cos x}\right)}

    =exp\left(\lim_{x\rightarrow 0}\frac{\ln\left(\frac{2-\cos x}{1-\cos x}\right)}{\frac{1}{\sin^{2}x}}\right)

    Use the L'Hospital's rule to evaluate the limit. Hope you can do it from here.
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  4. #4
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    Re: limit

    Quote Originally Posted by dhiab View Post
    Calculat :

    Click image for larger version. 

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    You could also use

    \lim_{x \to \infty}\left(1+\frac{1}{x}\right)^x=e

    while

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




    \lim_{x \to 0}\left(\frac{2-cosx}{1-cosx}\right)^{sin^2x}=\lim_{x \to 0}\left(\frac{1+1-cosx}{1-cosx}\right)^{1-cos^2x}

    =\lim_{x \to 0}\left[\left[\left(1+\frac{1}{1-cosx}\right)^{1-cosx}\right]^{1+cosx}\right]

    =\lim_{x \to 0}\left[\left[\lim_{y \to 0}\left(1+\frac{1}{y}\right)^{y}\right]^{1+cosx}\right]


    EDIT:

    A glaring error in the limit definition for "e"
    and some sloppiness in further lines have been edited.
    Thanks to CaptainBlack and Opalg.
    Last edited by Archie Meade; July 9th 2011 at 03:49 PM. Reason: terrible typos
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  5. #5
    Grand Panjandrum
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    Re: limit

    Quote Originally Posted by Archie Meade View Post
    You could also use

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

    Hence

    \lim_{x \to 0}\left(\frac{2-\cos x}{1-\cos x}\right)^{\sin^2x}=\lim_{x \to 0}\left(\frac{1+1-\cos x}{1-\cos x}\right)^{1-\cos^2x}

    \color{red}=\left[\lim_{x \to 0}\left(1+\frac{1}{1-\cos x}\right)^{1-\cos x}\right]^{1+\cos x}

    =\lim_{x \to 0}\left[e^{1-\cos x}\right]^{1+\cos x}

    =\lim_{x \to 0}e^{1-\cos^2x}=\lim_{x \to 0}e^{\sin^2x}
    You need to be more careful about line 5 (and onwards).

    CB
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  6. #6
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    Re: limit

    Quote Originally Posted by CaptainBlack View Post
    Quote Originally Posted by Archie Meade View Post
    You could also use

    \color{red}e^x=\lim_{x \to 0}\left(1+\frac{1}{x}\right)^x

    Hence

    \lim_{x \to 0}\left(\frac{2-cosx}{1-cosx}\right)^{sin^2x}=\lim_{x \to 0}\left(\frac{1+1-cosx}{1-cosx}\right)^{1-cos^2x}

    =\left[\lim_{x \to 0}\left(1+\frac{1}{1-cosx}\right)^{1-cosx}\right]^{1+cosx}

    =\lim_{x \to 0}\left[e^{1-cosx}\right]^{1+cosx}

    =\lim_{x \to 0}e^{1-cos^2x}=\lim_{x \to 0}e^{sin^2x}
    You need to be more careful about line 5 (and onwards).

    CB
    ... actually needs to be more careful from line 1 onwards!
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  7. #7
    Grand Panjandrum
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    Re: limit

    Quote Originally Posted by Archie Meade View Post
    You could also use

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

    Hence

    \lim_{x \to 0}\left(\frac{2-cosx}{1-cosx}\right)^{sin^2x}=\lim_{x \to 0}\left(\frac{1+1-cosx}{1-cosx}\right)^{1-cos^2x}

    =\left[\lim_{x \to 0}\left(1+\frac{1}{1-cosx}\right)^{1-cosx}\right]^{1+cosx}

    =\lim_{x \to 0}\left[e^{1-cosx}\right]^{1+cosx}

    =\lim_{x \to 0}e^{1-cos^2x}=\lim_{x \to 0}e^{sin^2x}
    OK what can we salvage from this?

    As (you may feel the need to prove this if you don't know it already):

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

    We may argue:

     \begin{array}{lcl}\displaystyle \lim_{x \to 0}\left(\frac{2-\cos(x)}{1-\cos(x)}\right)^{\sin^2(x)}&=&\displaystyle\lim_{x \to 0}\left(\frac{1+1-\cos(x)}{1-\cos(x)}\right)^{1-\cos^2(x)}\\ \\ &=&\displaystyle \lim_{x \to 0}\left\{ \left[ \left(1+\frac{1}{1-\cos(x)}\right)^{1-\cos(x)}\right]^{1+\cos(x)}  \right\}\\ \\&=&\displaystyle \left\{\lim_{x \to 0} \left[ \left(1+\frac{1}{1-\cos(x)}\right)^{1-\cos(x)}\right]\right\}^{\displaystyle\lim_{x\to 0}(1+\cos(x))}\\ \\&=&1^2=1  \end{array}

    Though a little care is required to justify the second from last line above (as always, it does help to know what the limit should be).

    CB
    Last edited by CaptainBlack; July 9th 2011 at 09:43 AM.
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