Question

Lt x tends to pi/6 (2-sqrt(3) cosx - sinx)/(6x-pi)2

Answer 1

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Answer 2

The limit of the given function is

$\lim_{x \to \frac{\pi }{6} \1} \frac{2-\sqrt{3}cosx-sinx }{(6x-\pi) ^2} = \frac{1}{36}$

• Solving the given limit, we have

         $\lim_{x \to \frac{\pi }{6} \1} \frac{2-\sqrt{3}cosx-sinx }{(6x-\pi) ^2}$

      Taking 2 common from the numerator, we get

        $\lim_{x \to \frac{\pi }{6} \1} \frac{2(1-\frac{\sqrt{3}}{2}cosx-\frac{1}{2} sinx) }{(6x-\pi) ^2}$      - (1)

• Now, we know that

               $cos\frac{\pi }{6} = \frac{\sqrt{3} }{2} \\and sin\frac{\pi }{6} = \frac{ 1}{2}$

    therefore (1) can be written as

        $=\lim_{x \to \frac{\pi }{6} \1} \frac{2(1-cosxcos\frac{\pi }{6} - sinxsin\frac{\pi }{6}) }{(6x-\pi) ^2}\\=\lim_{x \to \frac{\pi }{6} \1} \frac{2(1-(cosxcos\frac{\pi }{6}+ sinxsin\frac{\pi }{6})) }{(6x-\pi) ^2}$

• Now, Using the identity

       $cos(A-B) =cosAcosB + sinA sinB$

       we get,

      $=\lim_{x \to \frac{\pi }{6} \1} \frac{2(1-cos(x-\frac{\pi }{6} )) }{(6x-\pi) ^2}$

      $=2\lim_{x \to \frac{\pi }{6} \1} \frac{1-cos(x-\frac{\pi }{6} )}{(6x-\pi) ^2}$

• Using the the double angle formula for cosine that is

         $cos2A = 1 - 2sin^2A\\1 - cos2A= 2sin^2A$

        we get,

         $=2\lim_{x \to \frac{\pi }{6} \1} \frac{ 2sin^2(\frac{6x-\pi }{12} )}{(6x-\pi) ^2}$

         $=4\lim_{x \to \frac{\pi }{6} \1} \frac{ sin^2(\frac{6x-\pi }{12} )}{144(\frac{6x-\pi }{12})^2}$

• Now, using standard sine limit that is,

          $\lim_{x \to 0\1} \frac{sinx}{x} = 1$, we get

          $=4\lim_{x \to \frac{\pi }{6} \1} \frac{ sin^2(\frac{6x-\pi }{12} )}{144(\frac{6x-\pi }{12})^2}=4( \frac{1}{144})\\=\frac{1}{36}$

           Therefore we get,

             $\lim_{x \to \frac{\pi }{6} \1} \frac{2-\sqrt{3}cosx-sinx }{(6x-\pi) ^2} = \frac{1}{36}$