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Math Help - Limit

  1. #1
    Member kjchauhan's Avatar
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    Limit

    Please help me to prove this:

    \lim_{x\to\alpha} \dfrac{\alpha \sin x - x \sin \alpha} {\alpha \cos x - x \cos \alpha} = \tan (\alpha - \tan^{-1} \alpha)

    Hint is :

    \sin \alpha = \alpha - \frac {\alpha^3}{3!} + ...
    and
    \cos \alpha = 1 - \frac{\alpha^2}{2!} + ...

    Thanks in advance...
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  2. #2
    MHF Contributor Amer's Avatar
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    I did not used the hint

    \displaystyle \lim _{x \rightarrow a} \frac{a \sin x - x \cos a}{a\cos x - x \cos a}

    use lopital rule

    \displaystyle \lim _{x\rightarrow a} \frac{a \cos x - \cos a}{-a\sin x - \cos a}= \frac{ a \cos a - \sin a}{-a\sin a - \cos a}

    i want to show that \displaystyle \tan ( a - \tan ^{-1} a)

    \tan ( a - \tan ^{-1} a ) = \frac{\sin (a - \tan ^{-1} a)}{\cos (a - \tan^{-1} a)}= \frac{ a \cos a - \sin a}{-a\sin a - \cos a}

    remark that
    tan ^{-1} a = \sin ^{-1} \left( \frac{a}{\sqrt{a^2+1}}\right) = \cos ^{-1} \left( \frac{1}{\sqrt{a^2+1}} \right)

    now
    \displaystyle  \frac{\sin (a - \tan ^{-1} a)}{\cos (a - \tan^{-1} a)} = \frac{\sin a\cdot \cos (\tan ^{-1} a) - \cos a \cdot \sin (\tan ^{-1} a) }{\cos a \cdot \cos (\tan ^{-1} a ) + \sin a \sin ( \tan ^{-1} a) }


    simplify u will get \frac{a \cos a - \sin a}{-a\sin a - \cos a}

    or from beginning

    \displaystyle \tan(a - \tan ^{-1} a )  = \frac{\tan a - a }{1 +a \tan a} multiplying with cos a

    \displaystyle \frac{\sin a - a \cos a}{\cos a + a\sin a } and this is the limit result after using lopital
    Last edited by Amer; October 5th 2010 at 11:27 AM.
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  3. #3
    Member kjchauhan's Avatar
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    thanks.
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