It's obvious that, for a finite constant $c^*$ and a sequence $\{X_n : n \in \mathbb{N}\}$, the following two cannot be both true.
$ \lim\limits_{n \to \infty}{X_n} = +\infty$
$ \lim\limits_{n \to \infty}{X_n} = c^*$
REVISION & EDIT:
In repsponse to Karl's comment, we can suppose $X_n(\omega)$ is a function from a sample space, $f:\Omega \to X$. With the usual probability space {$\Omega, E, P$}, the limit notation $\lim\limits_{n}{X_n} = a$ is equivalent to saying that
$$ \lim\limits_{n}X_{n}(\omega)=a, \forall \omega \in \Omega \\ \iff Pr(\lim\limits_{n}X_{n}=a ) = 1 \wedge \nexists \{ M_0 \in E \ | \ Pr( \lim\limits_{n}X_{n}=b) = 0, X_n:M_0 \to X \} $$
where $a \ne b$.
With this clarification, the answers to following questions now seem to be obvious by themselves.
My questions are as followings.
Question 1:
Can the following two be both true?
$ \lim\limits_{n \to \infty}X_{n}(\omega) = +\infty, \forall \omega \in \Omega $
$ Prob( \lim\limits_{n \to \infty}{X_n} = c^* ) = 1$
Or vice versa, can the following two be both true?
$ \lim\limits_{n \to \infty}X_{n}(\omega) = c^*, \forall \omega \in \Omega $
$ Prob( \lim\limits_{n \to \infty}{X_n} = +\infty ) = 1$
And why or why not?
Question 2 (the weaker version):
Can the following two be both true?
$ \lim\limits_{n \to \infty}X_{n}(\omega) = +\infty, \forall \omega \in \Omega $
$ Prob( \lim\limits_{n \to \infty}{X_n} = c^* ) = 0$
Or vice versa, can the following two be both true?
$ \lim\limits_{n \to \infty}X_{n}(\omega) = c^*, \forall \omega \in \Omega $
$ Prob( \lim\limits_{n \to \infty}{X_n} = +\infty ) = 0$
Why or why not?