mathop {lim }limits_{x to 0} frac{{sin^2 x}}{ | vedclass

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Question

(B) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sin^2 x}}{{\sqrt{2} - \sqrt{1 + \cos x}}}$.Rationalizing the denominator:$L = \mathop {\lim }\li

Vedclass · Accepted Answer

$ 4\sqrt{2} $

Vedclass · Answer

(B) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sin^2 x}}{{\sqrt{2} - \sqrt{1 + \cos x}}}$.Rationalizing the denominator:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sin^2 x (\sqrt{2} + \sqrt{1 + \cos x})}}{{2 - (1 + \cos x)}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sin^2 x (\sqrt{2} + \sqrt{1 + \cos x})}}{{1 - \cos x}}$Using the identity $1 - \cos x = 2\sin^2(x/2)$ and $\sin x = 2\sin(x/2)\cos(x/2)$:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{(2\sin(x/2)\cos(x/2))^2 (\sqrt{2} + \sqrt{1 + \cos x})}}{{2\sin^2(x/2)}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{{4\sin^2(x/2)\cos^2(x/2) (\sqrt{2} + \sqrt{1 + \cos x})}}{{2\sin^2(x/2)}}$$L = \mathop {\lim }\limits_{x 	o 0} 2\cos^2(x/2) (\sqrt{2} + \sqrt{1 + \cos x})$As $x 	o 0$,$\cos(x/2) 	o 1$ and $\cos x 	o 1$:$L = 2(1)^2 (\sqrt{2} + \sqrt{1 + 1}) = 2(\sqrt{2} + \sqrt{2}) = 2(2\sqrt{2}) = 4\sqrt{2}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin^2 x}}{{\sqrt{2} - \sqrt{1 + \cos x}}}$ equals

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