Q73 / 150 · MHT CET 2022 (05 Aug Shift 1) · Mathematics

$\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$ equals

  • A
    $\sqrt{2}$
  • Correct answer
    B
    $4 \sqrt{2}$
  • C
    $2 \sqrt{2}$
  • D
    $4$

Solution

$\begin{aligned} & \lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}=\lim _{x \rightarrow 0} \frac{\sin ^2 x\{\sqrt{2}+\sqrt{1+\cos x}\}}{2-(1+\cos x)} \\ & =\lim _{x \rightarrow 0} \frac{(1-\cos x)(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})}{1-\cos x} \\ & =(1+\cos 0)(\sqrt{2}+\sqrt{1+\cos 0}) \\ & =2 \times 2 \sqrt{2} \\ & =4 \sqrt{2}\end{aligned}$

Hence, the correct answer is option: 'B'.

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