\mathop \lim \limits_x \to 0 \sin ^2x \over \sqrt 2 - \sqrt 1 + \cos x equals:

Limits Continuity and Differentiability PYQ — JEE Main 2019 · 8 Apr · Shift 1 · Q34 with Solution

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Correct answer: D

We need to evaluate $L=\lim_{x\to 0}\frac{\sin^2 x}{\sqrt{2}-\sqrt{1+\cos x}}.$
As $x\to 0$ , $\sin^2 x\to 0, \qquad \sqrt{2}-\sqrt{1+\cos x}\to \sqrt{2}-\sqrt{2}=0,$ so this is a $\frac{0}{0}$ form. We rationalize the denominator.
Multiply numerator and denominator by the conjugate: $L=\lim_{x\to 0}\frac{\sin^2 x}{\sqrt{2}-\sqrt{1+\cos x}}\cdot \frac{\sqrt{2}+\sqrt{1+\cos x}}{\sqrt{2}+\sqrt{1+\cos x}}.$

Then, $L=\lim_{x\to 0}\frac{\sin^2 x\left(\sqrt{2}+\sqrt{1+\cos x}\right)}{2-(1+\cos x)}.$

Simplify the denominator: $2-(1+\cos x)=1-\cos x.$ So, $L=\lim_{x\to 0}\frac{\sin^2 x\left(\sqrt{2}+\sqrt{1+\cos x}\right)}{1-\cos x}.$
Use the identity $\sin^2 x=(1-\cos x)(1+\cos x).$ Substituting, $L=\lim_{x\to 0}(1+\cos x)\left(\sqrt{2}+\sqrt{1+\cos x}\right).$
Now directly substitute $x=0$ : $1+\cos 0=1+1=2,$ and $\sqrt{2}+\sqrt{1+\cos 0}=\sqrt{2}+\sqrt{2}=2\sqrt{2}.$ Hence, $L=2\cdot 2\sqrt{2}=4\sqrt{2}.$
Therefore, the correct option is $\boxed{D\;:\;4\sqrt{2}}.$