lim _{x rightarrow 0} frac{sin ^2 x}{sqrt{2}- | vedclass

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Question

(B) We evaluate the limit: $\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$Rationalizing the denominator:$= \lim _{x \rightarrow 0

Vedclass · Accepted Answer

$4 \sqrt{2}$

Vedclass · Answer

(B) We evaluate the limit: $\lim _{x ightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$Rationalizing the denominator:$= \lim _{x ightarrow 0} \frac{\sin ^2 x (\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)}$$= \lim _{x ightarrow 0} \frac{\sin ^2 x (\sqrt{2}+\sqrt{1+\cos x})}{1-\cos x}$Using the identity $\sin ^2 x = 1-\cos ^2 x = (1-\cos x)(1+\cos x)$:$= \lim _{x ightarrow 0} \frac{(1-\cos x)(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})}{1-\cos x}$$= \lim _{x ightarrow 0} (1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})$Substituting $x = 0$:$= (1+\cos 0)(\sqrt{2}+\sqrt{1+\cos 0})$$= (1+1)(\sqrt{2}+\sqrt{1+1})$$= 2 	imes (\sqrt{2}+\sqrt{2})$$= 2 	imes 2 \sqrt{2} = 4 \sqrt{2}$

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

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