Answer
$\lim\limits_{n \to \infty} \frac{3^{n+1}+3}{3^n} =3$
Work Step by Step
$\lim\limits_{n \to \infty} \frac{3^{n+1}+3}{3^n} = \lim\limits_{n \to \infty} \frac{\frac{3^{n+1}}{3^n}+\frac{3}{3^n}}{\frac{3^n}{3^n}} =\lim\limits_{n \to \infty} \frac{3+\frac{3}{3^n}}{1} = \frac{3+0}{1} = 3$