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7.5 I · Q65

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

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Let t=π/6−xt=\pi/6-x, so x=π/6−tx=\pi/6-t and t→0t\to0; also π−6x=6t\pi-6x=6t. Expanding, cos⁡(π/6−t)=32cos⁡t+12sin⁡t\cos(\pi/6-t)=\tfrac{\sqrt3}2\cos t+\tfrac12\sin t and 3sin⁡(π/6−t)=32cos⁡t−32sin⁡t\sqrt3\sin(\pi/6-t)=\tfrac{\sqrt3}2\cos t-\tfrac32\sin t. Subtracting: cos⁡x−3sin⁡x=2sin⁡t\cos x-\sqrt3\sin x=2\sin t. So the e …

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