Answer
$\lim\limits_{n \to \infty}\frac{2e^n + 1}{e^n} = 2$
Work Step by Step
$\lim\limits_{n \to \infty}\frac{2e^n + 1}{e^n} = \lim\limits_{n \to \infty}\frac{\frac{2e^n}{e^n}+ \frac{1}{e^n}}{\frac{e^n}{e^n}} = \lim\limits_{n \to \infty}\frac{2+ \frac{1}{e^n}}{1} = \lim\limits_{n \to \infty}\frac{2+0}{1} = 2$
