Limit Inferior of Sequence $x_n = 2^n$

Question

I am doing a problem which said that the limit inferior of sequence $x_n = 2^n$ does not exist. But isn't that the limit inferior of this sequence is $2$, or there are different definitions of limit inferior? Thanks!

Answer 1

The limit inferior is defined as follows

$$\lim_{n \to \infty} \inf x_n := \lim_{n \to \infty }\big(\inf_{m\ge n }\hspace{4pt} x_m\big)$$

here the limit inferior is divergent thus does not exit.

Answer 2

$\beta_m=\inf\{x_n:n\ge m\}$. $\{\beta_m\}=\{2^1,2^2,...\}.$ We can easily prove that $\beta_m$ is divergent. Hence, Limit inferior doesn't exist.