Boundedness
Key Questions

Answer:
Boundedness is about having finite limits. In the context of values of functions, we say that a function has an upper bound if the value does not exceed a certain upper limit. More...
Explanation:
Other terms used are "bounded above" or "bounded below".
For example, the function
#f(x) = 1/(1+x^2)# is bounded above by#1# and below by#0# in that:#0 < f(x) <= 1# for all#x in RR# graph{1/(1+x^2) [5, 5, 2.5, 2.5]}
The function
#exp:x > e^x# is bounded below by#0# (or you can say has#0# as a lower bound), but is not bounded above.#0 < e^x < oo# for all#x in RR# graph{e^x [5.194, 4.806, 0.74, 4.26]}

Answer:
A continuous function defined on a closed interval has an upper (and lower) bound.
Explanation:
Probably the simplest boundedness theorem states that a continuous function defined on a closed interval has an upper (and lower) bound.
Proof by contradiction
Suppose
#f(x)# is defined and continuous on a closed interval#[a, b]# , but has no upper bound.Then:
#AA n in NN, EE x_n in [a, b] : f(x_n) > n# Since the sequence of
#x_n# 's lies in a bounded interval, it is dense at some point in the closure of the interval. Since the interval is closed, that must be at some point#c# actually in the interval#[a, b]# .Since the sequence of
#x_n# 's is dense at#c# , there is some monotonically increasing sequence#n_k in NN# such that#x_(n_k) > c# as#k>oo# .Now
#f(x)# is continuous at#c# , so:#lim_(x>c) f(x) = f(c)# which is bounded.
But:
#lim_(k>oo) x_(n_k) = c" "# and#" "lim_(k>oo) f(x_(n_k)) = oo# is unbounded.
...contradiction.
So there is no such
#f(x)# lacking upper (or lower) bound. 
If the function is unbounded, the graph would progress to infinity, in some direction(s).

Not all functions are bounded.
The simplest counter example would be
the identity function#f(x) = x#
which is defined for all values of#x# and can generate any value for#f(x)# A slightly less trivial counter example would be
the cubing function#f(x) = x^3#