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Math Help - Limit & Derivative Sin x..

  1. #1
    mmr
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    Limit & Derivative Sin x..

    1.
    \lim_{x\to 0}(\frac{\sin{x}}{x})^\frac{1}{x}
    2.
    \lim_{x\to 0}(\frac{\sin{3x}}{3x})^\frac{1}{x}

    Should I use just L'Hopital rules or Taylor's teorem?
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  2. #2
    Eater of Worlds
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    Quote Originally Posted by mmr View Post
    1.
    \lim_{x\to 0}(\frac{\sin{x}}{x})^\frac{1}{x}
    I assume you know that \lim_{x\to 0}\frac{sin(x)}{x}=1
    Last edited by mr fantastic; November 22nd 2008 at 04:43 PM. Reason: Added [/quote] to close off the quote.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mmr View Post
    1.
    \lim_{x\to 0}(\frac{\sin{x}}{x})^\frac{1}{x}
    2.
    \lim_{x\to 0}(\frac{\sin{3x}}{3x})^\frac{1}{x}

    Should I use just L'Hopital rules or Taylor's teorem?
    You could take the natural log and use taylors theorem. Or you could just use Maclaurin series \frac{\sin(x)}{x}=1-\frac{x^2}{3!}+\cdots\sim{1}
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  4. #4
    mmr
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    I assume you know that \lim_{x\to 0}\frac{sin(x)}{x}=1
    Yes but what about the \frac{1}{x} ?

    Quote Originally Posted by Mathstud28 View Post
    Just note that \sin(x)^{\frac{1}{x}}\to{1} and so does \frac{1}{x^{\frac{1}{x}}}
    Or you could take the natural log and use taylors theorem. Or you could just use Maclaurin series \frac{\sin(x)}{x}=1-\frac{x^2}{3!}+\cdots\sim{1}
    Ok, don't know Maclaurin series. To simplify the question, could I done it just with L'Hopital rules?
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mmr View Post
    Yes but what about the \frac{1}{x} ?



    Ok, don't know Maclaurin series. To simplify the question, could I done it just with L'Hopital rules?
    Yes, just take the natural log first.
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  6. #6
    mmr
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    \lim_{x\to 0}(\frac{\sin{x}}{x})^\frac{1}{x}=e^{\lim_{x\to 0}\frac{(ln{\frac{\sin{x}}{x}})'}{(x)'}}=e^{\lim_{  x\to 0}\frac{x\cos{x}-\sin{x}}{x\sin{x}}}
    Is this good? if yes,what next?
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mmr View Post
    \lim_{x\to 0}(\frac{\sin{x}}{x})^\frac{1}{x}=e^{\lim_{x\to 0}\frac{(ln{\frac{\sin{x}}{x}})'}{(x)'}}=e^{\lim_{  x\to 0}\frac{x\cos{x}-\sin{x}}{x\sin{x}}}
    Is this good? if yes,what next?
    Consider this, as x\to{0} we have that \sin(x)\to{x}

    So let us rewrite this limit as follows

    \begin{aligned}\lim_{x\to{0}}\frac{x\cos(x)-x}{x\sin(x)}&=\lim_{x\to{0}}\frac{x\cos(x)-x}{x^2}\\<br />
&=\lim_{x\to{0}}\frac{1-\cos(x)}{x}<br />
\end{aligned}

    Now this is a simple application of L'hopital's,
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