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Math Help - Triogo-limit

  1. #1
    Super Member dhiab's Avatar
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    Triogo-limit

    Calculate :
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  2. #2
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    Quote Originally Posted by dhiab View Post
    Calculate :
    Since this \to \frac{0}{0} you can use L'Hospital's Rule...

    \lim_{x \to 1}\frac{\sin{\left(\frac{\pi}{2x}\right)} - 1}{x^3 - 1} = \lim_{x \to 1}\frac{-\frac{\pi}{2x^2}\cos{\left(\frac{\pi}{2x}\right)}}  {3x^2}

     = \frac{-\frac{\pi}{2(1)^2}\cos{\left[\frac{\pi}{2(1)}\right]}}{3(1)^2}

    = 0.
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