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Math Help - limit

  1. #1
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    limit

    Hello everyone, please help me with this problem.
    \lim _{x\to 0}\frac{x}{\sin ^3 x}-\frac{\sin x}{x^3}.

    I will appreciate any help. Thank you very much.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: limit

    Using \sin x\sim x\;(x\to 0) we can express \dfrac{x}{\sin^3 x}-\dfrac{\sin^3 x}{x^3 }\sim \dfrac{x^4-\sin^6 x}{ x^6}
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  3. #3
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    Re: limit

    Thank you for your quick response, but my problem is

    \dfrac{x}{\sin^3 x}-\dfrac{\sin x}{x^3}

    , not -\dfrac{\sin ^3 x}{x^3}.
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  4. #4
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    Re: limit

    Quote Originally Posted by joll View Post
    Thank you for your quick response, but my problem is

    \dfrac{x}{\sin^3 x}-\dfrac{\sin x}{x^3}

    , not -\dfrac{\sin ^3 x}{x^3}.

    \frac{x}{(\sin(x))^3}-\frac{\sin(x)}{x^3}=\frac{x^4-(\sin(x))^4}{(\sin(x))^3x^3}

    Which is indeterminate so we can do a number of things, we could use L'Hopital's rule (twice?) or expand top and bottom as power series about x=0 or ...

    However we do this we get a finite limit.

    (P.S. you are allowed to use a calculator or some other mechanical assistance to get some idea of what the limit might be, which provides a useful cross check on your analysis)

    (P.P.S. I'm sure that the above quoted text is not what it said when I started answering this question!)

    CB
    Last edited by CaptainBlack; July 26th 2011 at 12:11 AM.
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  5. #5
    Grand Panjandrum
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    Re: limit

    Quote Originally Posted by FernandoRevilla View Post
    Using \sin x\sim x\;(x\to 0) we can express \dfrac{x}{\sin^3 x}-\dfrac{\sin^3 x}{x^3 }\sim \dfrac{x^4-\sin^6 x}{ x^6}

    \dfrac{x}{\sin^3 x}-\dfrac{\sin x}{x^3 }\sim \dfrac{x^4-\sin^4 x}{ x^6}

    CB
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Re: limit

    Quote Originally Posted by CaptainBlack View Post
    \dfrac{x}{\sin^3 x}-\dfrac{\sin x}{x^3 }\sim \dfrac{x^4-\sin^4 x}{ x^6}
    Perhaps, I read another thing. Now we can express \dfrac{x^4-\sin^4 x}{ x^6}=\dfrac{x^2+\sin^2 x}{x^3}\cdot \dfrac{x^2-\sin^2 x}{x^3} and the limit is easier to find.

    (P.P.S. I'm sure that the above quoted text is not what it said when I started answering this question!)
    I have the same impression.
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  7. #7
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    Re: limit

    Thank you for your quick response.
    Actually, I've tried L'Hospital's rule, but not twice. I will try that way first, thank you very much.
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  8. #8
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    Re: limit

    Thank you very much. That idea never ocurred to me; I will work in that way too.

    ----
    about the p.p.s ;

    I am terribly sorry for the confusion; actually, I didn't realize the mistake, and after I posted my reply saying

    \dfrac{x^4-(\sin x)^6}{x^6} = \dfrac{1}{x^2} - \left( \dfrac{\sin x}{x} \right)^6 \to \infty,

    I realized it was mistaken and totally editted my post. Perhaps I should have posted another reply. I am sorry.
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  9. #9
    MHF Contributor FernandoRevilla's Avatar
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    Re: limit

    Quote Originally Posted by joll View Post
    I realized it was mistaken and totally editted my post. Perhaps I should have posted another reply. I am sorry.
    It does not matter.
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