lim _{x rightarrow infty}left(frac{2+sin x}{x | vedclass

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(A) We know that for any real number $x$,the value of $\sin x$ lies in the interval $[-1, 1]$.Therefore,$2 + \sin x$ lies in the interval $[2 - 1, 2 +

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$0$

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(A) We know that for any real number $x$,the value of $\sin x$ lies in the interval $[-1, 1]$.Therefore,$2 + \sin x$ lies in the interval $[2 - 1, 2 + 1] = [1, 3]$.As $x ightarrow \infty$,the denominator $x^2 + 3 ightarrow \infty$.Thus,we have $\lim _{x}$ ${ightarrow \infty}\left(\frac{2+\sin x}{x^2+3}ight) = \frac{	ext{finite value between } 1 	ext{ and } 3}{\infty} = 0$.

$\lim _{x \rightarrow \infty}\left(\frac{2+\sin x}{x^2+3}\right)$ is equal to

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