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7.5 II · Q69

Q.lim⁡x→π/6[2−3cos⁡x−sin⁡x(6x−π)2]\displaystyle\lim_{x\to \pi/6}\left[\frac{2-\sqrt3\cos x-\sin x}{(6x-\pi)^2}\right]

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Let t=x−π/6t=x-\pi/6, so x=π/6+tx=\pi/6+t, t→0t\to0, and 6x−π=6t6x-\pi=6t. Using the angle-addition formulas, 3cos⁡(π/6+t)+sin⁡(π/6+t)=2cos⁡t\sqrt3\cos(\pi/6+t)+\sin(\pi/6+t)=2\cos t. So 2−3cos⁡x−sin⁡x=2(1−cos⁡t)=4sin⁡2(t/2)2-\sqrt3\cos x-\sin x=2(1-\cos t)=4\sin^2(t/2). Dividing by (6t)2=36t2(6t)^2=36t^2: $\dfrac{4\sin^2(t/2)}{36t^2}=\dfrac19\left(\d …

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