$\begin{align}  & \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,gof(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,g(\frac{x+1}{x-2})=\underset{x\to {{(-\frac{1}{2})}^{+}}}{\mathop{\lim }}\,g(x) \\  & =\underset{x\to {{(-\frac{1}{2})}^{+}}}{\mathop{\lim }}\,\frac{5x+1}{4{{x}^{2}}-1}=\frac{-\frac{3}{2}}{{{0}^{-}}}=+\infty  \\ \end{align}$