mathop {lim }limits_{x to 0} frac{{xcos x - s | vedclass

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(B) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{x\cos x - \sin x}}{{{x^2}\sin x}}$.Since the form is $\frac{0}{0}$ as $x \

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$-\frac{1}{3}$

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(B) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{x\cos x - \sin x}}{{{x^2}\sin x}}$.Since the form is $\frac{0}{0}$ as $x 	o 0$,we apply $L'	ext{Hospital's rule}$.First derivative of numerator: $\frac{d}{dx}(x\cos x - \sin x) = \cos x - x\sin x - \cos x = -x\sin x$.First derivative of denominator: $\frac{d}{dx}(x^2\sin x) = 2x\sin x + x^2\cos x$.So,the limit becomes $\mathop {\lim }\limits_{x 	o 0} \frac{-x\sin x}{2x\sin x + x^2\cos x} = \mathop {\lim }\limits_{x 	o 0} \frac{-\sin x}{2\sin x + x\cos x}$.Applying $L'	ext{Hospital's rule}$ again (still $\frac{0}{0}$ form):Numerator derivative: $\frac{d}{dx}(-\sin x) = -\cos x$.Denominator derivative: $\frac{d}{dx}(2\sin x + x\cos x) = 2\cos x + \cos x - x\sin x = 3\cos x - x\sin x$.Evaluating the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{-\cos x}{3\cos x - x\sin x} = \frac{-\cos(0)}{3\cos(0) - 0\sin(0)} = \frac{-1}{3(1) - 0} = -\frac{1}{3}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{x\cos x - \sin x}}{{{x^2}\sin x}} = $

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