## Thomas' Calculus 13th Edition

$1$
Consider the Taylor series for $\sin x$ as follows: $\sin x= x-\dfrac{ x^3}{3!}+\dfrac{x^5}{5!}-....$ $\lim\limits_{x \to \infty}(x+1) \sin (\dfrac{1}{x+1})=\lim\limits_{x \to \infty}(x+1) \times [-\dfrac{1}{3!(x+1)^3}+...]\\=\lim\limits_{x \to \infty} (1)- \lim\limits_{x \to \infty} \dfrac{1}{3!(x+1)^2}+\lim\limits_{x \to \infty} \dfrac{1}{5!(x+1)^4}-...)$ So, $\lim\limits_{x \to \infty}(x+1) \sin (\dfrac{1}{x+1})=1$