Categories > fun-lim > Limits > $\underset{x\to -2}{\mathop{\lim }}\,\frac{{{\sin }^{-1}}(x+2)}{{{x}^{2}}+2x}$ is equal to

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Question: $\underset{x\to -2}{\mathop{\lim }}\,\frac{{{\sin }^{-1}}(x+2)}{{{x}^{2}}+2x}$ is equal to


1) 0
2) $\infty $
3) –1/2
4) None of these
Solution: Explanation: No Explanation
Limits

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