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If $\{x_{n}\}_{n \geq 1}$ is a sequence of real numbers such that $\lim _{n \rightarrow \infty} \frac{x_{n}}{n}=0.001$, then

(A) $\{x_{n}\}_{n \geq 1}$ is a bounded sequence

(B)$\{x_{n}\}_{n \geq 1}$ is an unbounded sequence

(C) $\{x_{n}\}_{n \geq 1}$ is a convergent sequence

(D) $\{x_{n}\}_{n \geq 1}$ is a monotonically decreasing sequence

If $\{x_{n}\}_{n \geq 1}$ was bounded, show that $\lim _{n \rightarrow \infty} \frac{x_{n}}{n}=0$, by sandwich theorem.

If $\{x_{n}\}_{n \geq 1}$ was convergent, show that $\lim _{n \rightarrow \infty} \frac{x_{n}}{n}=0$, by algebra of limits.

If $\{x_{n}\}_{n \geq 1}$ was motonotically decreasing and bounded below, then it would have been convergent by Monotone Convergence Theorem.

Let's consider if it is not below below, i.e. $\lim _{n \rightarrow \infty} {x_{n}} = -\infty $

- Take $ {x_{n}} = -n $
- Take $ {x_{n}} = -n^2 $
- Take $ {x_{n}} = - \sqrt{n} $

Find the limit in each of this case.

Hence, it will be unbounded. See the full solution and proof idea below.

- What if, $\lim _{n \rightarrow \infty} \frac{x_{n}}{n}= 0$?
- Find examples.

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