lean_workbook_plus_43717

One way is like mcrasher shows: you can multiply $\frac{e^{x}}{4+5e^{3x}}$ by $\frac {e^{-x}}{e^{-x}}$ so that you get $\lim_{x\to\infty}\frac{1}{4e^{-x}+5e^{2x}}$ ( $e^{a}e^{b}=e^{a+b}$ ).