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Math Help - Trigo- limit( tan and sin)

  1. #1
    Super Member dhiab's Avatar
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    Trigo- limit( tan and sin)

    Calculate :


    \mathop {\lim }\limits_{x \to 0} \left( {\frac{{x - \tan \left( x \right)}}{{x - \sin \left( x \right)}}} \right)<br />
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    Quote Originally Posted by dhiab View Post
    Calculate :


    \mathop {\lim }\limits_{x \to 0} \left( {\frac{{x - \tan \left( x \right)}}{{x - \sin \left( x \right)}}} \right)<br />
    use L'Hopital's rule ... more than once in this case.

    you should get the limit to be -2
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by dhiab View Post
    Calculate :


    \mathop {\lim }\limits_{x \to 0} \left( {\frac{{x - \tan \left( x \right)}}{{x - \sin \left( x \right)}}} \right)<br />
    If you evaluate it, you end up with the indeterminate case \frac{0}{0}.

    Now, you can apply L'H˘pital's rule to get

    \lim_{x\to0}\frac{1-\sec^2x}{1-\cos x}=\lim_{x\to0}\frac{-\left(\sec^2x-1\right)}{1-\cos x}=\lim_{x\to0}\frac{-(\sec x+1)(\sec x-1)}{1-\cos x} =\lim_{x\to0}\frac{-(\sec x+1)(1-\cos x)}{\cos x(1-\cos x)}=\lim_{x\to0}-\frac{\sec x+1}{\cos x}=-2

    Does this make sense?
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