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Limits: Limiting Value

JEE Maths question with a full step-by-step solution.

Question
The limiting value of f(x)=22−(cos⁡x+sin⁡x)31−sin⁡2xf(x)=\dfrac{2\sqrt{2}-(\cos x+\sin x)^{3}}{1-\sin 2x} when x→π4x\to\dfrac{\pi}{4} is
A2\sqrt{2}
B12\dfrac{1}{\sqrt{2}}
C323\sqrt{2}
D32\dfrac{3}{\sqrt{2}}
Solution
Step 1: Substitute x=π4+hx=\dfrac{\pi}{4}+h where h→0h\to 0: cos⁡x+sin⁡x=2sin⁡(x+π4)=2sin⁡(π2+h)=2cos⁡h\cos x+\sin x=\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)=\sqrt{2}\sin\left(\dfrac{\pi}{2}+h\right)=\sqrt{2}\cos h So (cos⁡x+sin⁡x)3=22cos⁡3h(\cos x+\sin x)^{3}=2\sqrt{2}\cos^{3}h. Step 2: 1−sin⁡2x=1−sin⁡(π2+2h)=1−cos⁡2h1-\sin 2x=1-\sin\left(\dfrac{\pi}{2}+2h\right)=1-\cos 2h. Step 3: Substitute:
lim⁡h→022−22cos⁡3h1−cos⁡2h=lim⁡h→022(1−cos⁡3h)1−cos⁡2h\lim_{h\to 0}\dfrac{2\sqrt{2}-2\sqrt{2}\cos^{3}h}{1-\cos 2h}=\lim_{h\to 0}\dfrac{2\sqrt{2}(1-\cos^{3}h)}{1-\cos 2h}
 Step 4: Use 1−cos⁡2h=2sin⁡2h1-\cos 2h=2\sin^{2}h and factor 1−cos⁡3h=(1−cos⁡h)(1+cos⁡h+cos⁡2h)1-\cos^{3}h=(1-\cos h)(1+\cos h+\cos^{2}h):
=lim⁡h→022(1−cos⁡h)(1+cos⁡h+cos⁡2h)2sin⁡2h=\lim_{h\to 0}\dfrac{2\sqrt{2}(1-\cos h)(1+\cos h+\cos^{2}h)}{2\sin^{2}h}
 Step 5: Use 1−cos⁡h=2sin⁡2h21-\cos h=2\sin^{2}\dfrac{h}{2} and sin⁡2h=4sin⁡2h2cos⁡2h2\sin^{2}h=4\sin^{2}\dfrac{h}{2}\cos^{2}\dfrac{h}{2}:
=lim⁡h→022⋅2sin⁡2h2⋅(1+cos⁡h+cos⁡2h)2⋅4sin⁡2h2cos⁡2h2=\lim_{h\to 0}\dfrac{2\sqrt{2}\cdot 2\sin^{2}\dfrac{h}{2}\cdot(1+\cos h+\cos^{2}h)}{2\cdot 4\sin^{2}\dfrac{h}{2}\cos^{2}\dfrac{h}{2}}
=lim⁡h→022(1+cos⁡h+cos⁡2h)4cos⁡2h2=22⋅34⋅1=624=322=32=\lim_{h\to 0}\dfrac{2\sqrt{2}(1+\cos h+\cos^{2}h)}{4\cos^{2}\dfrac{h}{2}}=\dfrac{2\sqrt{2}\cdot 3}{4\cdot 1}=\dfrac{6\sqrt{2}}{4}=\dfrac{3\sqrt{2}}{2}=\dfrac{3}{\sqrt{2}}
 Correct answer: (4) 32\dfrac{3}{\sqrt{2}}
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