## Thomas' Calculus 13th Edition

converges to $1$.
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (8)^{\frac{1}{n}}$ as we know that $\lim\limits_{n \to \infty} x^{\frac{1}{n}}=1$ when $x \gt 0$ This shows that the limit has an indeterminate form so we will use L-Hospital's rule. This implies that $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (8)^{\frac{1}{n}}=1$ Thus, $\lim\limits_{n \to \infty} a_n=1$ converges to $1$.