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7.5 II · Q67

Q.lim⁡x→π/6[2sin⁡x−1π−6x]\displaystyle\lim_{x\to \pi/6}\left[\frac{2\sin x-1}{\pi-6x}\right]

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✓ Free question

Let t=π/6−xt=\pi/6-x, so x=π/6−tx=\pi/6-t, t→0t\to0, and π−6x=6t\pi-6x=6t. 2sin⁡(π/6−t)−1=2[12cos⁡t−32sin⁡t]−1=cos⁡t−3sin⁡t−1=(cos⁡t−1)−3sin⁡t2\sin(\pi/6-t)-1=2\left[\tfrac12\cos t-\tfrac{\sqrt3}2\sin t\right]-1=\cos t-\sqrt3\sin t-1=(\cos t-1)-\sqrt3\sin t. As t→0t\to0, (cos⁡t−1)/t→0(\cos t-1)/t\to0 while sin⁡t/t→1\sin t/t\to1, so dividing by 6t6t: (cos⁡t−1)−3sin⁡t6t→0−3(1)6=−36\dfrac{(\cos t-1)-\sqrt3\sin t}{6t}\to\dfrac{0-\sqrt3(1)}{6}=-\dfrac{\sqrt3}{6}.

✓Final answer

−36-\dfrac{\sqrt3}{6}

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