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Math Help - Limit computation

  1. #1
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    Limit computation

    Hi,
    How can i solve this next limit?

    lim (cos(x/√n))^n as n->infinity

    Thanks for any help...
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by rebecca View Post
    Hi,
    How can i solve this next limit?

    lim (cos(x/√n))^n as n->infinity

    Thanks for any help...
    \lim\limits_{n\to\infty}\cos^n\left(\frac{x}{\sqrt  {n}}\right)\to 1^{\infty}, an indeterminate form.

    What you'll need to do at this stage is take natural logs of both sides, and then apply L'Hopitals rule.

    Can you proceed?
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    I will continue from the point where mr. chris stopped...

    \lim_{\n\to\infty}cos^n(\frac{x}{\sqrt{n}})=e^\lim  _{\n\to\infty}lncos^n(\frac{x}{\sqrt{n}})
    (the last pass is from continuous of function e)

    Now, lets concentrate only in power of e.

    lim_{\n\to\infty}lncos^n(\frac{x}{\sqrt{n}})=lim_{  \n\to\infty}nlncos(\frac{x}{\sqrt{n}})

    And now we do a substitution: n=\frac{1}{t}.

    lim_{t\to 0}\frac{lncos({x}\sqrt{t})}{t}
    So we got a \frac{0}{0} form.

    Hence we can use l'Hôpital's rule.

    lim_{t\to 0}\frac{-xsin(x\sqrt(t))}{2cos(x\sqrt(t))\sqrt(t)}



    So we got again a \frac{0}{0} form.


    Hence we can use again l'Hôpital's rule.

    lim_{t\to 0}\frac{\frac{-x^2cos(x)\sqrt{t}}{2\sqrt{t}}}{\frac{-(xsin(x\sqrt(t))\sqrt(t)-cos(x\sqrt(t)))}{\sqrt{t}}}=\frac{-x^2}{2}


    So:

    \lim_{\n\to\infty}cos^n(\frac{x}{\sqrt{n}})=e^{\li  m_{\n\to\infty}lncos^n(\frac{x}{\sqrt{n}})}=e^{\fr  ac{-x^2}{2}}
    Last edited by Also sprach Zarathustra; June 24th 2010 at 11:45 AM.
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