Q103 / 160 · AP EAMCET 2021 (25 Aug Shift 2) · Mathematics

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

  • Correct answer
    A
    0
  • B
    1
  • C
    -1
  • D
    $\infty$

Solution

$\lim _{x \rightarrow \infty}\left(\frac{2+\sin x}{x^2+3}\right)$ Since, $\sin x \in[-1,1]$ Hence, when $x \rightarrow \infty$, then the function is of $\frac{\text { finite }}{\infty}$ form $\begin{aligned} & \lim _{x \rightarrow \infty}\left(\frac{2+\sin x}{x^2+3}\right) \\ &= \frac{2+\text { value in between }(-1) \text { and }(1)}{\infty+3}=0\end{aligned}$

Hence, the correct answer is option: 'A'.

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