I need show cases 1 and 2

Question

Prove that :

If lim inf$s_n$ = lim sup$s_n$ , then lim $s_n$ is defined and lim $s_n$ = lim inf$s_n$ = lim sup$s_n$

case(1) lim inf$s_n$ = lim sup$s_n$ = +∞ then lim $s_n$ = +∞.

case (2) if lim inf$s_n$ = lim sup$s_n$ = −∞ then lim $s_n$ = −∞. case(3)

case(3) Suppose that lim inf $s_n$ = lim sup $s_n$ = $s$ where $s$ is a real number. We need to prove lim $s_n$ = $s$. Let $ε > 0$ Since $s$ = lim vN there exists a positive integer $N_0$ such that $|s−sup{s_n :n>N_0}|$<ε. Thus sup${s_n :n>N_0}$<$s+ε$ , so $s_n$ <$s+ε$ for all n>$N_0$. (1) Similarly, there exists $N_1$ such that $|s − inf {s_n : n > N_1 }| $ < ε, hence inf${s_n:n>N_1}$ >$s−ε$, hence $s_n$ >$s−ε$ forall $n>N_1$. (2) From (1) and (2) we conclude $s−ε$<$s_n$<$s+ε $ for n>max${N_0,N_1}$ , equivalently $|s_n − s| $ < $ε$ for n > max${N_0,N_1}$. This proves lim $s_n$ = $s$ as desired.

Answer 1

Hint

Indefinite limits can be vaguely defined using a phrase "For however large $K$ there exists $\varepsilon$ such that any values of $s_n$ are higher than $K$.", hence no need to consider supremum for case (1) and infimum for case (2).

I hope I did not spoil the fun playing around with the $\varepsilon$'s, it is a great exercise.