Question: Let $(a\_n)$ positive sequence where $$\dfrac{n-1}{n} \leq \dfrac{a\_{n+1}}{a\_n} \leq \dfrac {n}{n+1}$$

Prove that: $\lim\_{n\to\infty} a\_n = 0$.

I was stuck here:

I: $$\lim\_{n\to\infty} \dfrac{n-1}{n} = \lim\_{n\to\infty}1-\dfrac{1}{n} = 1$$
II: $$\lim\_{n\to\infty} \dfrac {n}{n+1} = \lim\_{n\to\infty} \dfrac{1}{1+ \dfrac{1}{n}} = 1$$

Then: $$\lim\_{n\to\infty} \dfrac{a\_{n+1}}{a\_n} = 1$$

How do I continue from here?

Answer: 
Just observe that $$a\_{n+1}\le \frac{n}{n+1}a\_n\le \frac{n}{n+1}\cdot\frac{n-1}{n}a\_{n-1}=\frac{n-1}{n+1}a\_{n-1}\le\cdots\le \frac{1}{n}a\_1\\a\_{n+1}\ge \frac{n-1}{n}a\_n\ge \frac{n-1}{n}\cdot\frac{n-2}{n-1}a\_{n-1}=\frac{n-2}{n}a\_{n-1}\ge\cdots\ge \frac{1}{n}a\_2\\\implies \lim\sup\_n a\_n\le 0\le \lim\inf\_n a\_n\\\implies \lim\_n a\_n=0$$ Actually the lower bound on $a\_n/a\_{n-1}$ is redundant as $a\_n$ is a positive sequence.
@Samrat Mukhopadhyay Another solution, in the same vein as yours:

Let us define: $b\_n:=na\_n$.

The second inequality can thus be written: $b\_{n+1}\leq b\_n.$ 

Thus $b\_n$ is a decreasing sequence of positive numbers.

Being bounded from below by zero, sequence $b\_n$ converges to its "lim inf", say $M \geq 0$. 

From $\lim\_{n \to \infty} b\_n = M$ we conclude that, using the definition of $b\_n$,

$$\lim\_{n \to \infty} a\_n = \lim\_{n \to \infty} b\_n \dfrac{1}{n} = M\times 0 = 0.$$ 

Remark: surprisingly, the first inequality doesn't play any role.