Answer
$1$
Work Step by Step
Taylor series for $\sin x $ can be defined as: $\sin x= x-\dfrac{ x^3}{3!}+\dfrac{x^5}{5!}-....$
Now, $\lim\limits_{x \to \infty}(x+1) \sin (\dfrac{1}{x+1})=\lim\limits_{x \to \infty}(x+1) (-\dfrac{1}{3!(x+1)^3}+...)$
or, $=\lim\limits_{x \to \infty}(1-\dfrac{1}{3!(x+1)^2}+\dfrac{1}{5!(x+1)^4}-...)$
or, $=1$
