mathop {lim }limits_{x to 0} frac{{sqrt {1 + | vedclass

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(B) To evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}{x}$,we can use $L'Hospital's$ rule since

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(B) To evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}{x}$,we can use $L'Hospital's$ rule since it is a $\frac{0}{0}$ indeterminate form.Differentiating the numerator and denominator with respect to $x$:Numerator: $\frac{d}{dx}(\sqrt{1+\sin x} - \sqrt{1-\sin x}) = \frac{\cos x}{2\sqrt{1+\sin x}} - \frac{-\cos x}{2\sqrt{1-\sin x}} = \frac{\cos x}{2\sqrt{1+\sin x}} + \frac{\cos x}{2\sqrt{1-\sin x}}$Denominator: $\frac{d}{dx}(x) = 1$Applying the limit as $x 	o 0$:$\mathop {\lim }\limits_{x 	o 0} \left( \frac{\cos x}{2\sqrt{1+\sin x}} + \frac{\cos x}{2\sqrt{1-\sin x}} ight) = \frac{\cos 0}{2\sqrt{1+\sin 0}} + \frac{\cos 0}{2\sqrt{1-\sin 0}}$$= \frac{1}{2\sqrt{1}} + \frac{1}{2\sqrt{1}} = \frac{1}{2} + \frac{1}{2} = 1.$

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}{x} = $

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