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Limits — JEE Maths practice question

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

Question

The limiting value of

f(x)=\dfrac{2\sqrt{2}-(\cos x+\sin x)^{3}}{1-\sin 2x}

when

x\to\dfrac{\pi}{4}

\sqrt{2}

\dfrac{1}{\sqrt{2}}

3\sqrt{2}

\dfrac{3}{\sqrt{2}}

Solution

Step 1: Substitute

x=\dfrac{\pi}{4}+h

where

h\to 0

\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

(\cos x+\sin x)^{3}=2\sqrt{2}\cos^{3}h

. Step 2:

1-\sin 2x=1-\sin\left(\dfrac{\pi}{2}+2h\right)=1-\cos 2h

. Step 3: Substitute:

\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=2\sin^{2}h

and factor

1-\cos^{3}h=(1-\cos h)(1+\cos h+\cos^{2}h)

=\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=2\sin^{2}\dfrac{h}{2}

and

\sin^{2}h=4\sin^{2}\dfrac{h}{2}\cos^{2}\dfrac{h}{2}

=\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\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)

\dfrac{3}{\sqrt{2}}

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