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Math Help - Trigo- limit

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

    Calculate :
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  2. #2
    MHF Contributor red_dog's Avatar
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    1) \lim_{x\to\frac{\pi}{4}}(1-\sin 2x)\frac{\sin 2x}{\cos 2x}=\lim_{x\to\frac{\pi}{4}}\frac{(\cos x-\sin x)^2\sin 2x}{\cos^2x-\sin^2x}=

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

    2) We have \cos 2x=2\cos^2x-1=2\left(\cos^2x-\frac{1}{2}\right)=

    =2\left(\cos^2x-\cos^2\frac{\pi}{4}\right)=2\left(\cos x-\cos\frac{\pi}{4}\right)\left(\cos x+\cos\frac{\pi}{4}\right)

    Then \lim_{x\to\frac{\pi}{4}}\frac{\cos 2x}{\cos\frac{\pi}{4}-\cos x}=

    =\lim_{x\to\frac{\pi}{4}}\frac{2\left(\cos x-\cos\frac{\pi}{4}\right)\left(\cos x+\cos\frac{\pi}{4}\right)}{\cos\frac{\pi}{4}-\cos x}=

    =-\lim_{x\to\frac{\pi}{4}}2\left(\cos x+\cos\frac{\pi}{4}\right)=-2\sqrt{2}

    Or you can use l'Hospital.
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  3. #3
    Super Member dhiab's Avatar
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    Hello : ANATHOR SOLUTION


    but :


    and :

    conclusion :

    Thank you
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  4. #4
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    Quote Originally Posted by dhiab View Post
    Calculate :
    L'Hospital's Rule would be easiest to use here...

    (1 - \sin{2x})\tan{2x} = \frac{\sin{2x} - \sin^2{2x}}{\cos{2x}}.

    As x \to \frac{\pi}{4}, the fraction \to \frac{0}{0}.


    Using L'Hospital, this limit is the same as

    \lim_{x \to \frac{\pi}{4}}\left(\frac{2\cos{2x} - 4\sin{2x}\cos{2x}}{-\sin{2x}}\right)

     = \frac{-0}{-1}

     = 0.
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  5. #5
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    Quote Originally Posted by dhiab View Post
    Calculate :
    As above, the easiest way to evaluate the second is with L'Hospital.

    The fraction \to \frac{0}{0}.

    So using L'Hospital, it is the same as

    \lim_{x \to \frac{\pi}{4}}\left(\frac{-2\sin{2x}}{\sin{x}}\right)

     = \frac{-2\sin{\frac{\pi}{2}}}{\sin{\frac{\pi}{4}}}

     = \frac{-2}{\frac{1}{\sqrt{2}}}

     = -2\sqrt{2}.
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