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$\lim _{x \rightarrow(-3)}\left(\frac{\sin ^{-1}(x+3)}{x^2+3 x}\right)$ is equal to
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$\frac{-1}{3}$
$\begin{aligned} & \lim _{x \rightarrow-3} \frac{\sin ^{-1}(x+3)}{x^2+3 x} \\ & \lim _{x \rightarrow-3} \frac{\sin ^{-1}(x+3)}{x(x+3)}=-\frac{1}{3}\end{aligned}$
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