Answer
The solution is
$$\lim_{x\to\infty}\frac{e^{3x}}{3e^{3x}+5}=\frac{1}{3}$$
Work Step by Step
To solve this limit follow the steps below
$$\lim_{x\to\infty}\frac{e^{3x}}{3e^{3x}+5}=\lim_{x\to\infty}\frac{e^{3x}}{e^{3x}\left(3+\frac{5}{e^{3x}}\right)}=\lim_{x\to\infty}\frac{1}{3+5e^{-3x}}=\left[\frac{1}{3+5e^{-3\cdot\infty}}\right]=\left[\frac{1}{3+0}\right]=\frac{1}{3}.$$
