mathop {lim }limits_{x to infty } frac{{sin x | vedclass

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(B) We need to evaluate $\mathop {\lim }\limits_{x 	o \infty } \frac{{\sin x}}{x}$.We know that for all $x \in \mathbb{R}$,the value of $\sin x$ is b

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(B) We need to evaluate $\mathop {\lim }\limits_{x 	o \infty } \frac{{\sin x}}{x}$.We know that for all $x \in \mathbb{R}$,the value of $\sin x$ is bounded such that $-1 \le \sin x \le 1$.For $x > 0$,we can divide the inequality by $x$:$-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}$.Applying the Squeeze Theorem as $x 	o \infty$:$\mathop {\lim }\limits_{x 	o \infty } \left( -\frac{1}{x} ight) = 0$ and $\mathop {\lim }\limits_{x 	o \infty } \left( \frac{1}{x} ight) = 0$.Therefore,by the Squeeze Theorem,$\mathop {\lim }\limits_{x 	o \infty } \frac{{\sin x}}{x} = 0$.

$\mathop {\lim }\limits_{x \to \infty } \frac{{\sin x}}{x} = $

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