Evaluate $\lim\limits_{n \to \infty} nx_n.$ Suppose $$x_2 \in \left(0,\frac{\pi}{2}\right), x_{n+1}=\left(1-\frac{1}{n}\right)\sin x_n(n \ge 2).$$ Evaluate $\lim\limits_{n \to \infty} nx_n.$
Note that $$x_{n+1}=\left(1-\frac{1}{n}\right)\sin x_n\le \sin x_n\le x_n,$$ and $$|x_{n+1}|=\left|\left(1-\frac{1}{n}\right)\sin x_n\right|\le 1,$$ it follows that $\{x_n\}$ converges. Therefore, we can readily obtain $x_n \to 0$.
As for the limit wanted, we can consider apply Stolz theorem. But it's too complicated.
 A: Work in progress. We can conclude that $(nx_n)_{n\in\mathbb{N}}$ converges ...

Using induction we can show that $x_n \geq 0$ and, thus, $nx_n\geq0$.
Function $\sin{x}\geq0$, for $x\in\left[0,\frac{\pi}{2}\right]$. As a result $\sin{x_2}>0$ and
$$0\leq x_3=\frac{\sin{x_2}}{2}\leq 1<\frac{\pi}{2} \Rightarrow x_3\in\left[0,\frac{\pi}{2}\right]$$
And using the induction hypothesis for $x_n\in \left[0,\frac{\pi}{2}\right]$
$$0\leq x_{n+1}=\left(1-\frac{1}{n}\right)\sin{x_n}\leq 1<\frac{\pi}{2} \Rightarrow x_{n+1}\in\left[0,\frac{\pi}{2}\right]$$

From
$$|x_{n+1}|=
\left|1-\frac{1}{n}\right||\sin{x_n}|\leq
\left|1-\frac{1}{n}\right||x_n|\leq \\
\left|1-\frac{1}{n}\right|\left|1-\frac{1}{n-1}\right||\sin{x_{n-1}}|\leq ... \\
\leq \left|1-\frac{1}{n}\right|\left|1-\frac{1}{n-1}\right|\left|1-\frac{1}{n-2}\right|...\left|1-\frac{1}{2}\right||\sin{x_2}|=\frac{|\sin{x_2}|}{n}$$
Thus
$$|(n+1)x_{n+1}|\leq \left(1+\frac{1}{n}\right)|\sin{x_2}|$$
and $(nx_n)_{n\in\mathbb{N}}$ is bounded.

Now:
$$0\leq \frac{(n+1)x_{n+1}}{nx_{n}}=\left(1+\frac{1}{n}\right)\left(1-\frac{1}{n}\right) \frac{\sin{x_n}}{x_n} \leq 1-\frac{1}{n^2}<1$$
and $(nx_n)_{n\in\mathbb{N}}$ is descending.
