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Q. $\displaystyle\lim_{x\to0} \frac{\sin^{2}x}{\sqrt{2} - \sqrt{1+\cos x}} $ equals :

JEE MainJEE Main 2019Limits and Derivatives

Solution:

$\lim_{x\to0} \frac{\left(\frac{\sin^{2}x}{x^{2}}\right)\left(\sqrt{2} + \sqrt{1+\cos x}\right)}{\left(\frac{1-\cos x}{x^{2}}\right)}$
$ = \frac{\left(1\right)^{2} .\left(2\sqrt{2}\right)}{\frac{1}{2}} = 4\sqrt{2}$