mathop {lim }limits_{x to 0} ,left( {frac{{x | vedclass

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Question

(B) Let $L = \mathop {\lim }\limits_{x 	o 0} \left( {\frac{{x - \sin x}}{x}} ight)\sin \left( {\frac{1}{x}} ight)$.We can rewrite the expression

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equals $0$

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(B) Let $L = \mathop {\lim }\limits_{x 	o 0} \left( {\frac{{x - \sin x}}{x}} ight)\sin \left( {\frac{1}{x}} ight)$.We can rewrite the expression as:$L = \mathop {\lim }\limits_{x 	o 0} \left( {1 - \frac{\sin x}{x}} ight) \sin \left( {\frac{1}{x}} ight)$.We know that $\mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{x} = 1$,so $\mathop {\lim }\limits_{x 	o 0} \left( {1 - \frac{\sin x}{x}} ight) = 1 - 1 = 0$.The term $\sin \left( {\frac{1}{x}} ight)$ oscillates between $-1$ and $1$ as $x 	o 0$,meaning it is a bounded function.By the Squeeze Theorem,since the limit of the first part is $0$ and the second part is bounded,the product is $0 	imes (	ext{bounded value}) = 0$.Therefore,the limit is $0$.

$\mathop {\lim }\limits_{x \to 0} \,\left( {\frac{{x - \sin x}}{x}} \right)\,\sin \left( {\frac{1}{x}} \right)$

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