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7.5 II · Q71

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

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Let t=x−π/2t=x-\pi/2, so x=π/2+tx=\pi/2+t, t→0t\to0, and 2x−π=2t2x-\pi=2t. cos⁡3x=cos⁡(3π/2+3t)=sin⁡3t\cos3x=\cos(3\pi/2+3t)=\sin3t (since cos⁡(3π/2)=0,sin⁡(3π/2)=−1\cos(3\pi/2)=0,\sin(3\pi/2)=-1). Also cos⁡x=cos⁡(π/2+t)=−sin⁡t\cos x=\cos(\pi/2+t)=-\sin t, so 3cos⁡x=−3sin⁡t3\cos x=-3\sin t. The numerator is sin⁡3t−3sin⁡t\sin3t-3\sin t; using the triple-angle identity sin⁡3t=3sin⁡t−4sin⁡3t\sin3t=3\sin t-4\sin^3t, this equals −4sin⁡3t-4\sin^3t. So the expression is $\dfr …

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