Is it possible to replace $\lim$ by $\sup$ here? Let $(a_m)$ be a monotonically increasing sequence. Is it then possible that $\lim_{m\to\infty} a_m=\sup_m a_m$?
I only know the fact that a bounded monotonically increasing sequence coverges to its supremum. But in the question above, the sequence is not supposed to be bounded.
 A: You hit the nail on the head.  Since the sequence is not necessarily bounded, we don't know that it has a least upper bound, i.e., a sup.  According to the definition, supremum only makes sense on bounded sequences because it has to satisfy two things:  It is a real number that's larger than every element of the set, but smaller than every other real number bounding the set.
Since there is no real number bounding the set, we can't talk about the smallest real number bounding the set.
A: Two cases are possible: Either the sequence is bounded and in that case you know that the equation holds.
Or the sequence is not bounded. In that case, the sequence diverges and in fact tends to infinity, usually written as $\lim_{m\to\infty}a_m=\infty$. Also, the supremum of an unbounded set is usually denoted as $\infty$, so that with this convention (i.e., in the extended real) the given equality holds in all cases. 
A: *

*if the sequence is not bounded: $\sup a_m = \infty$; and as the sequence is increasing and not bounded, its limit is $\infty$. Hence $\lim a_m = \sup a_m$.

*otherwise: the sequence is convergent. As the sequence is increasing:
$$
a_n \le \lim a_m \implies \sup a_m \le \lim a_m \\
a_n \le \sup a_m \implies \lim a_m \le \sup a_m
$$
the first inequality is because of the definition of the sup bound;
the second is a property of the limit.
