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Math Help - Limit of a function with sine

  1. #1
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    Limit of a function with sine

    Hello,

    I am trying to calculate the following limit:

    \lim\limits_{x\rightarrow+\infty}{\frac{1}{x-\sqrt{x^2+1}}\sin{\frac{1}{x}}}

    So far I've multiplied by \frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2+1}}, which yields the following:

    \lim\limits_{x\rightarrow+\infty}{\frac{x+\sqrt{x^  2+1}}{-1}\sin{\frac{1}{x}}} = \lim\limits_{x\rightarrow+\infty}{-(x+\sqrt{x^2+1})\sin{\frac{1}{x}}}

    I don't really know how to continue from here. I'd appreciate any help.
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  2. #2
    MHF Contributor alexmahone's Avatar
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  3. #3
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    Quote Originally Posted by thomasdotnet View Post
    I am trying to calculate the following limit:

    \lim\limits_{x\rightarrow+\infty}{\frac{1}{x-\sqrt{x^2+1}}\sin{\frac{1}{x}}}

    So far I've multiplied by \frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2+1}}, which yields the following:

    \lim\limits_{x\rightarrow+\infty}{\frac{x+\sqrt{x^  2+1}}{-1}\sin{\frac{1}{x}}} = \lim\limits_{x\rightarrow+\infty}{-(x+\sqrt{x^2+1})\sin{\frac{1}{x}}}
    First, \lim_{x\to\infty}x\sin\tfrac1x = \lim_{x\to\infty}\frac{\sin\frac1x}{\frac1x} = \lim_{y\to0}\frac{\sin y}y = 1 (where y=1/x). Then

    \lim_{x\to+\infty}-(x+\sqrt{x^2+1})\sin\tfrac1x = \lim_{x\to+\infty}-(1+\sqrt{1+x^{-2}})x\sin\tfrac1x = -2.
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  4. #4
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    That solves my problem. Thank you.
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