[Math] Lim Sup Calculation: $\limsup_{n\to\infty} b^n$

Question

I want to calculate $\lim \sup b^n$, where $b$ is some nonnegative number.

Now, if $0 \le b \le 1$, I expect the lim sup to be $0$, because these numbers are getting tinier and tinier, and we are taking the supremum of all these tiny numbers approaching zero. When $b > 1$, I expect lim sup to be infinity. However, I haven't done many calculations with lim sups, so I am having difficulty.

Note, I am permitted to assume any ideas coming from a course on real analysis, so we don't necessarily have to use the original definition of lim sup.

EDIT:

Here is a proof of $\lim \inf b^n = 0$ when $0 \le b < 1$.

If $0 \le b \le 1$, then $0 \le b^n \le 1$, and we know it is a bounded convergent sequence. Now, we have that

$0 \le \inf \{ b^k ~|~ k \ge n \} \le b^n \le 1$ for all $n \in \mathbb{N}$,

which means we can say something about the limit of these two sequences—namely, that

$\lim_{n \rightarrow \infty} \inf \{ b^k ~|~ k \ge n \} \le \lim_{n \rightarrow \infty} b^n$,

or

$\lim \inf b^n \le 0$,

which means $\lim \inf b^n = 0$.

Does this seem right?

Answer 1

Hint: $\limsup=\liminf=\lim$ if the limit exists.