## Calculus (3rd Edition)

$2$
Using Theorem one, we have $$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}{4+\frac{1}{n}}=4.$$ Now, since $\sqrt{x}$ is continuous for all $x\gt 0$, then by Theorem 4, we have $$\lim_{n\to \infty}\sqrt{4+\frac{1}{n}}=\sqrt{ \lim_{n\to \infty}(4+\frac{1}{n})}=\sqrt{4}=2.$$