ابتدا معادلهٔ توابع $f$ و $g$ را می‌یابیم. $\begin{align}  & moadeley\,tabee\,\,''f'':\,\frac{x}{3}+\frac{y}{2}=1\Rightarrow f(x)=\frac{-2}{3}x+2 \\  & moadeley\,tabee\,\,''g'':\,\frac{x}{2}+\frac{y}{-2}=1\Rightarrow g(x)=x-2 \\  & \underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{f(x)}{g(x)}=\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{-\frac{3}{2}x+2}{x-2}=\frac{-2}{3} \\ \end{align}$