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Math Help - lim n->infinity for cos function

  1. #1
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    lim n->infinity for cos function

    lim n-> infinity (cos(4*x/(n^.5)))^n

    I can plug this into mathematica and get e^-(8*x^2) but I don't know how that answer was obtained...
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by soma View Post
    lim n-> infinity (cos(4*x/(n^.5)))^n

    I can plug this into mathematica and get e^-(8*x^2) but I don't know how that answer was obtained...
    Note that \lim_{n\to\infty}\left(\cos\!\left(\frac{4x}{n^{.5  }}\right)\right)^{n}\to 1^{\infty}

    Let L=\lim_{n\to\infty}\left(\cos\!\left(\frac{4x}{n^{  .5}}\right)\right)^{n}. Applying natural log to both sides gives us

    \ln L=\lim_{n\to\infty}\frac{\ln\left(\cos\!\left(\fra  c{4x}{n^{.5}}\right)\right)}{\frac{1}{n}}=\frac{0}  {0}

    Now apply L'H˘pital's rule.

    Can you try to take it from here?
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  3. #3
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    For L'H˘pital's rule, I derived the top and bottom and got 0/0 again, and the second derivative looks daunting. Can you help me further with this?
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Note that \lim_{n\to\infty}\left(\cos\!\left(\frac{4x}{n^{.5  }}\right)\right)^{n}\to 1^{\infty}

    Let L=\lim_{n\to\infty}\left(\cos\!\left(\frac{4x}{n^{  .5}}\right)\right)^{n}. Applying natural log to both sides gives us

    \ln L=\lim_{n\to\infty}\frac{\ln\left(\cos\!\left(\fra  c{4x}{n^{.5}}\right)\right)}{\frac{1}{n}}=\frac{0}  {0}

    Now apply L'H˘pital's rule.

    Can you try to take it from here?
    Quote Originally Posted by soma View Post
    For L'H˘pital's rule, I derived the top and bottom and got 0/0 again, and the second derivative looks daunting. Can you help me further with this?
    After applying L'H˘pital's rule once, we have

    \lim_{n\to\infty}\frac{2x\tan\left(\frac{4x}{n^{.5  }}\right)}{-\frac{1}{n^{.5}}}=\frac{0}{0}

    Applying it another time, we see that

    \lim_{n\to\infty}-\frac{\frac{4x^2}{n^{1.5}}\sec^2\left(\frac{4x}{n^  {.5}}\right)}{\frac{1}{2n^{1.5}}}=\lim_{n\to\infty  }-8x^2\sec^2\left(\frac{4x}{n^{.5}}\right)=-8x^2

    Therefore, \ln L=-8x^2\implies L=\lim_{n\to\infty}\left(\cos\left(\frac{4x}{n^{.5  }}\right)\right)^n=e^{-8x^2}

    Does this make sense?

    (Note that when we differentiated, it was with respect to n, not x).
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  5. #5
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    Perfect!

    It took me a while to compute those derivatives myself, but I see how it all fits together now. Thanks so much!
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