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Q.Evaluate lim(x→π/4) (4√2-(cos x+sin x)^5)/(1-sin 2x).

West Bengal WbchseWest Bengal HS First Year (WBCHSE Class XI) Annual Examination 2018Subjective· 4mImportance★★★★★
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Substitute x=π/4+tx=\pi/4+t so that cos⁡x+sin⁡x=2cos⁡t\cos x+\sin x=\sqrt2\cos t and 1−sin⁡2x=2sin⁡2t1-\sin2x=2\sin^2t, then use the standard limit lim⁡t→01−cos⁡ntt2=n2\lim_{t\to0}\dfrac{1-\cos^nt}{t^2}=\dfrac n2.

Let x=π4+tx=\dfrac\pi4+t, so t→0t\to0 as x→π4x\to\dfrac\pi4.

cos⁡x+sin⁡x=2cos⁡ ⁣(x−π4)=2cos⁡t\cos x+\sin x=\sqrt2\cos\!\left(x-\dfrac\pi4\right)=\sqrt2\cos t (a standard identity), so

(cos⁡x+sin⁡x)5=(2)5cos⁡5t=42cos⁡5t.(\cos x+\sin x)^5=(\sqrt2)^5\cos^5t=4\sqrt2\cos^5t.

Numerator: 42−42cos⁡5t=42(1−cos⁡5t)4\sqrt2-4\sqrt2\cos^5t=4\sqrt2(1-\cos^5t).

1−sin⁡2x=1−sin⁡ ⁣(π2+2t)=1−cos⁡2t=2sin⁡2t1-\sin2x=1-\sin\!\left(\dfrac\pi2+2t\right)=1-\cos2t=2\sin^2t (using 1−cos⁡2t=2sin⁡2t1-\cos2t=2\sin^2t).

So the expression becomes

42(1−cos⁡5t)2sin⁡2t=22⋅1−cos⁡5tsin⁡2t.\dfrac{4\sqrt2(1-\cos^5t)}{2\sin^2t}=2\sqrt2\cdot\dfrac{1-\cos^5t}{\sin^2t}. …

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