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7.5 II · Q68

Q.lim⁡x→π/4[2−cos⁡x−sin⁡x(4x−π)2]\displaystyle\lim_{x\to \pi/4}\left[\frac{\sqrt2-\cos x-\sin x}{(4x-\pi)^2}\right]

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✓ Free question

Let t=x−π/4t=x-\pi/4, so x=π/4+tx=\pi/4+t, t→0t\to0, and 4x−π=4t4x-\pi=4t. Using the angle-addition formulas, cos⁡x+sin⁡x=cos⁡(π/4+t)+sin⁡(π/4+t)=2cos⁡t\cos x+\sin x=\cos(\pi/4+t)+\sin(\pi/4+t)=\sqrt2\cos t. So 2−cos⁡x−sin⁡x=2(1−cos⁡t)=22sin⁡2(t/2)\sqrt2-\cos x-\sin x=\sqrt2(1-\cos t)=2\sqrt2\sin^2(t/2). Dividing by (4t)2=16t2(4t)^2=16t^2: 22sin⁡2(t/2)16t2=28(sin⁡(t/2)t/2)214→28×1×14=232\dfrac{2\sqrt2\sin^2(t/2)}{16t^2}=\dfrac{\sqrt2}{8}\left(\dfrac{\sin(t/2)}{t/2}\right)^2\dfrac14\to\dfrac{\sqrt2}{8}\times1\times\dfrac14=\dfrac{\sqrt2}{32}.

✓Final answer

232\dfrac{\sqrt2}{32}

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