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Math Help - How to find these limits?

  1. #1
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    How to find these limits?

    Hello! Today was my first calculus exam, and I couldn't solve these to problems. Can you help me with these?

    \lim_{x\to\0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}

    \lim_{n\to\infty}n\cdot(\sqrt[n]{2011}-1)

    And is this convergent, why or why not?
    \sum_{k=1}^{\infty}\frac{1}{\sqrt{k}}\sin^2\frac{1  }{\sqrt{k}}

    Thank you, and happy holidays!
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  2. #2
    odx
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    Re: How to find these limits?

    \lim_{n \to \infty }n(\sqrt[n]{2011}-1)=\lim_{n \to \infty }\frac{2011^{\frac{1}{n}}-1}{\frac{1}{n}}=ln 2011
    Last edited by odx; December 23rd 2011 at 04:21 AM.
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  3. #3
    Grand Panjandrum
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    Re: How to find these limits?

    Quote Originally Posted by gotmejerry View Post
    And is this convergent, why or why not?
    \sum_{k=1}^{\infty}\frac{1}{\sqrt{k}}\sin^2\frac{1  }{\sqrt{k}}
    Since:

    \frac{1}{\sqrt{k}}\sin^2\left(\frac{1}{\sqrt{k}} \right)=O(k^{-3/2})

    the series converges.

    CB
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  4. #4
    Member sbhatnagar's Avatar
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    Re: How to find these limits?

    \lim_{x \to 0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}

    \frac{0}{0 } indeterminate form

    \begin{align*} \lim_{x \to 0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}} &= \lim_{x \to 0} \frac{x^{2 \over 3}\sin{x}}{\sin(\sqrt[3]{x})\cos(x^{2 \over 3})} \quad [\text{L 'Hospital's rule}]\\ \\ &= -\lim_{x \to 0} \frac{\sqrt[3]{n\cdot \cos{x}}[2\sin{x}+3x\cos{x}]}{2x^{2\over 3}\sin{\sqrt[3]{x}}\sin{x}-\cos{\sqrt[3]{x}}\cos{x}} \quad [\text{L 'Hospital's rule}] \\ \\ &= -\frac{0}{-1} \\ \\&=0\end{align*}
    Last edited by sbhatnagar; December 23rd 2011 at 05:27 AM.
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