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Math Help - Limit

  1. #1
    Junior Member
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    Limit

    Hi everybody,

    i must calculate these following limits:

    1)- \lim_{x \to 0} \frac{1-coxcos2x...cosnx}{x^2}

    2)- \lim_{x \to 0}\frac{1-cosxcos^2(2x)...cos^n(nx)}{x^2}

    Can you help me please?

    And thank you anyway.
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  2. #2
    MHF Contributor red_dog's Avatar
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    1) L=\lim_{x\to 0}\frac{1-\cos x\cos 2x\ldots\cos nx}{x^2}=

    (add and substract \cos x at numerator)

    =\lim_{x\to 0}\frac{1-\cos x+\cos x-\cos x\cos 2x\ldots\cos nx}{x^2}=

    =\lim_{x\to 0}\frac{1-\cos x}{x^2}+\lim_{x\to 0}\frac{\cos x(1-\cos 2x\cos 3x\ldots\cos nx)}{x^2}=

    =\lim_{x\to 0}\frac{1-\cos x}{x^2}+\lim_{x\to 0}\frac{1-\cos 2x\cos 3x\ldots\cos nx}{x^2}

    Do the same thing to the second limit: add and substract \cos 2x, and so on. Then

    L=\sum_{k=1}^n\lim_{x\to 0}\frac{1-\cos kx}{x^2}=\sum_{k=1}^n\lim_{x\to 0}\frac{2\sin^2\frac{kx}{2}}{x^2}=

    =2\sum_{k=1}^n\frac{k^2}{4}=\frac{1}{2}\sum_{k=1}^  n k^2=\frac{n(n+1)(2n+1)}{12}

    2) Using the same method we have

    L=\lim_{x\to 0}\frac{1-\cos x\cos^22x\ldots\cos^nnx}{x^2}=

    \lim_{x\to 0}\frac{1-\cos x}{x^2}+\lim_{x\to 0}\frac{1-\cos^22x}{x^2}+\lim_{x\to 0}\frac{1-\cos^33x}{x^2}+\ldots+\lim_{x\to 0}\frac{1-\cos^nnx}{x^2}=

    =\sum_{k=1}^n\lim_{x\to 0}\frac{1-\cos^kkx}{x^2}=

    =\sum_{k=1}^n\lim_{x\to 0}\frac{(1-\cos kx)(1+\cos kx+\cos^2kx+\ldots+\cos^{k-1}kx)}{x^2}=

    =\sum_{k=1}^n\lim_{x\to 0}\frac{2\sin^2\frac{kx}{2}}{x^2}\cdot\lim_{x\to 0}(1+\cos kx+\cos^2kx+\ldots\cos^{k-1}kx)=

    =\frac{1}{2}\sum_{k=1}^nk^3=\frac{1}{2}\left(\frac  {n(n+1)}{2}\right)^2=\frac{n^2(n+1)^2}{8}
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  3. #3
    Junior Member
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    Quote Originally Posted by red_dog View Post
    1) L=\lim_{x\to 0}\frac{1-\cos x\cos 2x\ldots\cos nx}{x^2}=

    (add and substract \cos x at numerator)

    =\lim_{x\to 0}\frac{1-\cos x+\cos x-\cos x\cos 2x\ldots\cos nx}{x^2}=

    =\lim_{x\to 0}\frac{1-\cos x}{x^2}+\lim_{x\to 0}\frac{\cos x(1-\cos 2x\cos 3x\ldots\cos nx)}{x^2}=

    =\lim_{x\to 0}\frac{1-\cos x}{x^2}+\lim_{x\to 0}\frac{1-\cos 2x\cos 3x\ldots\cos nx}{x^2}

    Do the same thing to the second limit: add and substract \cos 2x, and so on. Then

    L=\sum_{k=1}^n\lim_{x\to 0}\frac{1-\cos kx}{x^2}=\sum_{k=1}^n\lim_{x\to 0}\frac{2\sin^2\frac{kx}{2}}{x^2}=

    =2\sum_{k=1}^n\frac{k^2}{4}=\frac{1}{2}\sum_{k=1}^  n k^2=\frac{n(n+1)(2n+1)}{12}

    2) Using the same method we have

    L=\lim_{x\to 0}\frac{1-\cos x\cos^22x\ldots\cos^nnx}{x^2}=

    \lim_{x\to 0}\frac{1-\cos x}{x^2}+\lim_{x\to 0}\frac{1-\cos^22x}{x^2}+\lim_{x\to 0}\frac{1-\cos^33x}{x^2}+\ldots+\lim_{x\to 0}\frac{1-\cos^nnx}{x^2}=

    =\sum_{k=1}^n\lim_{x\to 0}\frac{1-\cos^kkx}{x^2}=

    =\sum_{k=1}^n\lim_{x\to 0}\frac{(1-\cos kx)(1+\cos kx+\cos^2kx+\ldots+\cos^{k-1}kx)}{x^2}=

    =\sum_{k=1}^n\lim_{x\to 0}\frac{2\sin^2\frac{kx}{2}}{x^2}\cdot\lim_{x\to 0}(1+\cos kx+\cos^2kx+\ldots\cos^{k-1}kx)=

    =\frac{1}{2}\sum_{k=1}^nk^3=\frac{1}{2}\left(\frac  {n(n+1)}{2}\right)^2=\frac{n^2(n+1)^2}{8}
    Thank you very much.
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