## Thomas' Calculus 13th Edition

converges to $1$
As we know that when $x \gt 0$ so, $\lim\limits_{n \to \infty} \sqrt[n] {n}=1$ and $\lim\limits_{n \to \infty} x^{1/n}=1$ Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (n+4)^{1/n+4}$ Consider $u=n+4$ So, $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (n+4)^{1/n+4}$ and $a_n=\lim\limits_{n \to \infty} (u)^{(1/u)}=1$ Thus, $\lim\limits_{n \to \infty} a_n=1$ and {$a_n$} converges to $1$