two limit questions related to $\sin$ I stuck with these 2 limits, can you help me please?
$1.\displaystyle\quad \lim_{n \to \infty }\frac{\sin1+2\sin\frac{1}{2}+3\sin\frac{1}{3}+\cdots+n\sin\frac{1}{n}}{n}$
$2.\displaystyle\quad \lim_{n \to \infty }\frac{n}{\frac{1}{\sin1}+\frac{1/2}{\sin1/2}+\frac{1/3}{\sin1/3}+\cdots+\frac{1/n}{\sin1/n}} 
$
Thanks in advance.
 A: We will use unnecessarily explicit inequalities to prove the result. 
In the first limit, the general term on top can be rewritten as $\dfrac{\sin(1/k)}{1/k}$. This reminds us of the $\frac{\sin x}{x}$ whose limit as $x\to 0$ we needed in beginning calculus. 
Note that for $0\lt x\le 1$, the power series 
$$x-\frac{x^3}{3!}+\frac{x^5}{5!} -\frac{x^7}{7!}+\cdots$$
for $\sin x$ is an alternating series. It follows that for $0\lt x\le 1$,
$$x-\frac{x^3}{6}\lt \sin x\lt x.$$
and therefore
$$1-\frac{x^2}{6}\lt \frac{\sin x}{x}\lt 1.$$
Put $x=1/k$. We get
$$1-\frac{1}{6k^2}\lt \frac{\sin(1/k)}{1/k} \lt 1.\tag{$1$}$$
Add up, $k=1$ to $k=n$, and divide by $n$ Recall that 
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots =\frac{\pi^2}{6}.$$
We find that
$$1-\frac{\pi^2}{36n}\lt \frac{\sin1+2\sin\frac{1}{2}+3\sin\frac{1}{3}+\cdots+n\sin\frac{1}{n}}{n}\lt 1.$$
From this, it follows immediately that our limit is $1$. 
A very similar argument works for the second limit that was asked about. It is convenient to consider instead the reciprocal, and calculate
$$\lim_{n \to \infty }\frac{\frac{1}{\sin1}+\frac{1/2}{\sin1/2}+\frac{1/3}{\sin1/3}+\cdots+\frac{1/n}{\sin1/n}}{n}.$$ 
We can then use the inequality
$$1\lt \frac{1/k}{\sin(1/k)} \lt \frac{1}{1-\frac{1}{6k^2}},$$
which is simple to obtain from the Inequalities $(1)$. 
Having the $1-\frac{1}{6k^2}$ in the denominator is inconvenient, so we can for example use the inequality $\dfrac{1}{1-\frac{1}{6k^2}}\lt 1+\dfrac{1}{k^2}$ to push through almost the same proof as the first one.
A: As $n \to \infty , \sin(1/n)\to 1/n , n\sin(1/n)\to 1$. Now, by Cesaro mean 
$\lim\limits_{n \to \infty} \sum_{1}^{n}n\sin(1/n)\to n$. 
Distributing the it over numerator and denominator
$\lim\limits_{n \to \infty} \frac{\sum_{1}^{n}n\sin(1/n)}{n}= \frac{\lim\limits_{n \to \infty }n\sin(1/n)}{n}=1$
So, the answer to the first part is 1 . Same argument holds for the second part too.
A: the first one , you can use the stolz theorem directly.
or use the result:
if 
$\lim\limits_{n\to \infty}a_n=a$, then $\lim\limits_{n\to \infty}\frac{a_1+a_2+.....a_n}{n}=a$, you can use the $\epsilon-N$ to illustrate it...
the second is the same 
