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Thread: limit-trigo

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

    calculate this limit ( no hospital ) :

    $\displaystyle
    \lim_{x\rightarrow \frac{\pi }{3}}(\frac{2cos(x)+1}{x-\frac{\pi }{3}})
    $
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by dhiab View Post
    calculate this limit ( no hospital ) :

    $\displaystyle
    \lim_{x\rightarrow \frac{\pi }{3}}(\frac{2cos(x)+1}{x-\frac{\pi }{3}})
    $
    Because:

    $\displaystyle \displaystyle \lim_{x\to \frac{\pi }{3}}\left[2\cos(x)+1\right] \ne 0$

    (since $\displaystyle \cos(\pi/3)=1/2$ ), The limit does not exist.

    Or are you asking for:

    $\displaystyle
    \displaystyle \lim_{x\to \frac{\pi }{3}}\left(\frac{2\cos(x)-1}{x-\frac{\pi }{3}}\right)
    $

    (Expanding the numerator as a Taylor series about $\displaystyle \pi/3$ this can be seen to be $\displaystyle -\sqrt{3}$ )


    CB
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