lim _{x rightarrow 0} frac{sqrt{1+x sin x}-sq | vedclass

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(D) The given limit is $\lim _{x ightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{	an ^2 2 x}$,which is of the $\frac{0}{0}$ form. Rationalizing

Vedclass · Accepted Answer

$\frac{3}{16}$

Vedclass · Answer

(D) The given limit is $\lim _{x ightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{	an ^2 2 x}$,which is of the $\frac{0}{0}$ form. Rationalizing the numerator: $= \lim _{x}$ ${ightarrow 0} \frac{(\sqrt{1+x \sin x}-\sqrt{\cos x})(\sqrt{1+x \sin x}+\sqrt{\cos x})}{	an ^2 2 x (\sqrt{1+x \sin x}+\sqrt{\cos x})}$ $= \lim _{x ightarrow 0} \frac{1+x \sin x-\cos x}{	an ^2 2 x} \cdot \frac{1}{\sqrt{1+x \sin x}+\sqrt{\cos x}} $ $= \lim _{x ightarrow 0} \frac{(1-\cos x)+x \sin x}{	an ^2 2 x} \cdot \frac{1}{2} $ Using the identities $1-\cos x = 2 \sin ^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$: $= \lim _{x}$ ${ightarrow 0} \frac{2 \sin ^2 \frac{x}{2} + 2x \sin \frac{x}{2} \cos \frac{x}{2}}{	an ^2 2 x} \cdot \frac{1}{2}$ $= \lim _{x}$ ${ightarrow 0} \frac{2 \sin ^2 \frac{x}{2} [1 + \frac{x \cos \frac{x}{2}}{\sin \frac{x}{2}}]}{	an ^2 2 x} \cdot \frac{1}{2}$ $= \lim _{x ightarrow 0} \frac{2 \sin ^2 \frac{x}{2} [1 + \frac{x}{	an \frac{x}{2}}]}{	an ^2 2 x} \cdot \frac{1}{2} $ $= \lim _{x ightarrow 0} \frac{2 (\frac{x}{2})^2 [1 + 2 \cdot \frac{x/2}{	an (x/2)}]}{(2x)^2} \cdot \frac{1}{2} $ $= \lim _{x}$ ${ightarrow 0} \frac{2 \cdot \frac{x^2}{4} [1+2]}{4x^2} \cdot \frac{1}{2} = \frac{1/2 \cdot 3}{4} \cdot \frac{1}{2} = \frac{3}{16}$.

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

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