Q66 / 170 · Manipal (MET) 2019 · Mathematics

$\lim _{x \rightarrow-2} \frac{\sin ^{-1}(x+2)}{x^2+2 x}$ is equal to

  • A
    $0$
  • B
    $\infty$
  • Correct answer
    C
    $-\frac{1}{2}$
  • D
    None of these

Solution

$\begin{aligned} & \text { We have, } y=\lim _{x \rightarrow-2} \frac{\sin ^{-1}(x+2)}{x^2+2 x} \\ & \begin{aligned} & \Rightarrow \quad y=\lim _{x \rightarrow-2} \frac{\sin ^{-1}(x+2)}{(x+2)} \times \frac{1}{x} \\ & \Rightarrow \quad y=\lim _{x \rightarrow-2}(1) \times \frac{1}{x} \quad\left[\because \lim _{x \rightarrow \theta} \frac{\sin \theta}{\theta}=1\right] \\ &=-\frac{1}{2}\end{aligned}\end{aligned}$

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

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