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Math Help - calculus limits

  1. #1
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    Post calculus limits

    Hi mathsforum

    I needed some help with the following:

    Find

    1. \lim_{x \to 0} \left\{ \frac{1-2sin^2x-cos^3x}{5x^2}\right\}


    2. \lim_{x \to 0} \left\{ \frac{x-sinx}{x-tanx}\right\}


    thank you.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by dadon View Post

    1. \lim_{x \to 0} \left\{ \frac{1-2sin^2x-cos^3x}{5x^2}\right\}
    This is of the form \frac{0}{0} so we may use L'Hopital's rule:
    \lim_{x \to 0}\frac{1-2sin^2(x)-cos^3(x)}{5x^2}

     = \lim_{x \to 0}\frac{-4sin(x)cos(x) + 3sin(x)cos^2(x)}{10x}

    This is still of the form \frac{0}{0} so we use L'Hopital's rule again:
     = \lim_{x \to 0}\frac{-4cos^2(x) + 4sin^2(x) + 3cos^3(x) - 3sin^2(x)cos(x)}{10} = \frac{-4 + 3}{10} = -\frac{1}{10}

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by dadon View Post
    2. \lim_{x \to 0} \left\{ \frac{x-sinx}{x-tanx}\right\}
    Again using L'Hopital's rule:
    \lim_{x \to 0}\frac{x-sinx}{x-tanx}

     = \lim_{x \to 0}\frac{1 - cos(x)}{-tan^2(x)}

    Again apply L'Hopital's rule:
     = \lim_{x \to 0}\frac{sin(x)}{-2tan^3(x) - 2tan(x)}

    Again:
     = \lim_{x \to 0}\frac{cos(x)}{-6tan^4(x) - 8tan^2(x) - 2} = \frac{1}{-2} = - \frac{1}{2}

    -Dan
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