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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits2 Marks
MCQ
$\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$ equals
  • A
    4
  • B
    $2 \sqrt{2}$
  • ✓
    $4 \sqrt{2}$
  • D
    $\sqrt{2}$
✓

Answer

Correct option: C.
$4 \sqrt{2}$
(C)
$\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$
$=\lim _{x \rightarrow 0}\left(\frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}\right)\left(\frac{\sqrt{2}+\sqrt{1+\cos x}}{\sqrt{2}-\sqrt{1+\cos x}}\right)$
$=\lim _{x \rightarrow 0}\left(\frac{\sin ^2 x}{1-\cos x}\right)(\sqrt{2}+\sqrt{1+\cos x})$
$=\lim _{x \rightarrow 0} \frac{\sin ^2 x}{2 \sin ^2\left(\frac{x}{2}\right)}(\sqrt{2}+\sqrt{1+\cos x})$
$=\lim _{x \rightarrow 0} \frac{1}{2}\left(\frac{\sin x}{x}\right)^2 \times x^2 \times\left(\frac{\frac{x}{2}}{\sin \frac{x}{2}}\right)^2$ $\times \frac{1}{\left(\frac{x^2}{4}\right)} \times(\sqrt{2}+\sqrt{1+\cos x})$
$=\frac{1}{2} \times 4 \times 2 \sqrt{2}=4 \sqrt{2}$

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