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Math Help - Evaluating limits

  1. #1
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    Evaluating limits

    lim x->0 (x cos x - sin x) / (x - sin x)
    lim x->0 (-x sin x) / (1 - cos x)
    lim x->0 (- sin x - x cos x) / (sin x)
    lim x->0 (-2cos x + x sin x) / (cos x)
    = -2

    This is what's written in my book (Numerical Analysis). It has a few review questions on limits but it doesn't go in to detail so I'm not sure how to go from step 1 to step 2.
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  2. #2
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    Quote Originally Posted by Makall View Post
    lim x->0 (x cos x - sin x) / (x - sin x)
    lim x->0 (-x sin x) / (1 - cos x)
    lim x->0 (- sin x - x cos x) / (sin x)
    lim x->0 (-2cos x + x sin x) / (cos x)
    = -2

    This is what's written in my book (Numerical Analysis). It has a few review questions on limits but it doesn't go in to detail so I'm not sure how to go from step 1 to step 2.
    1. Since this tends to \frac{0}{0}, you can use L'Hospital's Rule.

    \lim_{x \to 0} \frac{x\cos{x} - \sin{x}}{x - \sin{x}} = \lim_{x \to 0}\frac{\frac{d}{dx}(x\cos{x} - \sin{x})}{\frac{d}{dx}(x - \sin{x})}

     = \lim_{x \to 0}\frac{\cos{x} - x\sin{x} - \cos{x}}{1 - \cos{x}}

     = \lim_{x \to 0}\frac{-x\sin{x}}{1 - \cos{x}}


    This still tends to \frac{0}{0}, so use L'Hospital's Rule again.

     = \lim_{x \to 0}\frac{\frac{d}{dx}(-x\sin{x})}{\frac{d}{dx}(1 - \cos{x})}

     = \lim_{x \to 0}\frac{-\sin{x} - x\cos{x}}{\sin{x}}


    This still tends to \frac{0}{0}, so use L'Hospital's Rule again.

     = \lim_{x \to 0}\frac{\frac{d}{dx}(-\sin{x} - x\cos{x})}{\frac{d}{dx}(\sin{x})}

     = \lim_{x \to 0}\frac{-\cos{x} - \cos{x} + x\sin{x}}{\cos{x}}

     = \lim_{x \to 0}\frac{x\sin{x} - 2\cos{x}}{\cos{x}}

     = \frac{-2}{1}

     = -2.
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  3. #3
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    Quote Originally Posted by Makall View Post
    lim x->0 (x cos x - sin x) / (x - sin x)
    lim x->0 (-x sin x) / (1 - cos x)
    lim x->0 (- sin x - x cos x) / (sin x)
    lim x->0 (-2cos x + x sin x) / (cos x)
    = -2

    This is what's written in my book (Numerical Analysis). It has a few review questions on limits but it doesn't go in to detail so I'm not sure how to go from step 1 to step 2.
    2. \lim_{x \to 0}\frac{-x\sin{x}}{1 - \cos{x}}

    Since this tends to \frac{0}{0}, you can use L'Hospital's Rule.

     = \lim_{x \to 0}\frac{\frac{d}{dx}(-x\sin{x})}{\frac{d}{dx}(1 - \cos{x})}

     = \lim_{x \to 0}\frac{-\sin{x} - x\cos{x}}{\sin{x}}

    Since this tends to \frac{0}{0}, use L'Hospital's Rule again.

     = \lim_{x \to 0}\frac{\frac{d}{dx}(-\sin{x} - x\cos{x})}{\frac{d}{dx}(\sin{x})}

     = \lim_{x \to 0}\frac{-\cos{x} - \cos{x} + x\sin{x}}{\cos{x}}

     = \lim_{x \to 0}\frac{-2\cos{x} + x\sin{x}}{\cos{x}}

     = \frac{-2}{1}

     = -2.


    BTW I thought the second step was a whole new question - could someone delete this post please?
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