$X_n$ ($n\ge1)$ is sequence of real numbers such that $\lim_{n \to \infty}{\frac{x_n}{n}}=0.001$ then if $X_n$ ($n\ge1)$ is sequence of real numbers such that $\lim_{n \to \infty}{\frac{x_n}{n}}=0.001$ then,
$(A)\ \ X_n$ is a bounded sequence
$(B)\ \ X_n$ is a unbounded sequence
$(C)\ \ X_n$ is a Convvergent sequence
$(D)\ \ X_n$ is a monotonically decreasing sequence
I can only think of this particular case.
$\text{convergent sequence}\cdot \frac{1}{n}=\text{convergent sequence}\cdot \text{convergent sequence} =$finite limit
Therefore it is convergent but I think its not a correct way to solve this problem.
How do I figure out what is our sequence behavior just on the basis of limits provided?
 A: In the comments you considered the example $x_n = 0.001 n$ and that correctly eliminated choices A, C, D.  So B is the only remaining choice.
It is useful to know why B holds true, that is, why any sequence $\{x_n\}$ that satisfies $x_n/n\rightarrow 0.001$ must be unbounded. Knowing that it is true for a particular example is not good enough.  One way to explore this is through "proof by contradiction": Assume $\{x_n\}_{n=1}^{\infty}$ is bounded, so that there is a finite number $M$ such that
$$ -M \leq x_n \leq M \quad \forall n \in \{1, 2, 3, ...\}$$
Now reach a contradiction with the assumption $x_n/n\rightarrow 0.001$.
A: $(A)$ $\lim_{n \to \infty}{\frac{x_n}{n}} \leq \lim_{n \to \infty}{\frac{M}{n}} = 0 < 0.001$
$(C)$ $\lim_{n \to \infty}{\frac{x_n}{n}}=0.001 \implies 0=\lim_{n \to \infty}{\frac{L}{n}}=0.001$.
$(D)$ If $(x_{n})$ is eventually constant, we have the same case as $(C)$.
Otherwise, $(x_{n})$ is strictly decreasing and $(x_{n})$ is eventually always negative which implies that if $\lim_{n \to \infty}\frac{x_{n}}{n}$ exists, then $\lim_{n \to \infty}{\frac{x_n}{n}} < 0 < 0.001 $.
Then the answer must be $(B)$.
Finding an example (that satisfies the hypothesis but just so happens to not satisfy any of the other properties...) may also work as you have shown.
A: A direct proof.
If $\lim_{n \to \infty} x_n/n = c > 0$
then, for all large enough $n$,
$x_n/n > c/2$
so $x_n > cn/2$
and this is unbounded.
