## Thomas' Calculus 13th Edition

converges to $1$.
As we know that a sequence converges when $\lim\limits_{n \to \infty}a_n$ exists. Consider $a_n=1+(0.9)^n$ Apply limits to both sides. $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}[1+(0.9)^n]$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}(1)+\lim\limits_{n \to \infty}(0.9)^n$ $\lim\limits_{n \to \infty}a_n=1+0$ $\lim\limits_{n \to \infty}a_n=1$ Therefore, the sequence converges to $1$.