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7.4 Q.III · Q59

Q.lim⁡x→π[1−cos⁡x−2sin⁡2x]\displaystyle\lim_{x\to \pi}\left[\frac{\sqrt{1-\cos x}-\sqrt2}{\sin^2x}\right]

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✓ Free question

Let t=x−πt=x-\pi, so x=π+tx=\pi+t and t→0t\to0. Then cos⁡x=−cos⁡t\cos x=-\cos t, so 1−cos⁡x=1+cos⁡t=2cos⁡2(t/2)1-\cos x=1+\cos t=2\cos^2(t/2), giving 1−cos⁡x=2cos⁡(t/2)\sqrt{1-\cos x}=\sqrt2\cos(t/2) for small tt. So the numerator 1−cos⁡x−2=2[cos⁡(t/2)−1]=−22sin⁡2(t/4)\sqrt{1-\cos x}-\sqrt2=\sqrt2[\cos(t/2)-1]=-2\sqrt2\sin^2(t/4). Also sin⁡x=−sin⁡t\sin x=-\sin t, so sin⁡2x=sin⁡2t=4sin⁡2(t/2)cos⁡2(t/2)\sin^2x=\sin^2t=4\sin^2(t/2)\cos^2(t/2). The ratio is −22sin⁡2(t/4)4sin⁡2(t/2)cos⁡2(t/2)\dfrac{-2\sqrt2\sin^2(t/4)}{4\sin^2(t/2)\cos^2(t/2)}; using sin⁡2(t/4)/sin⁡2(t/2)→1/4\sin^2(t/4)/\sin^2(t/2)\to1/4 (from matching the small-angle ratios) and cos⁡2(t/2)→1\cos^2(t/2)\to1, this tends to −22×(1/4)4=−28\dfrac{-2\sqrt2\times(1/4)}{4}=-\dfrac{\sqrt2}{8}.

✓Final answer

−28-\dfrac{\sqrt2}{8}

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