Limit of a Monotonic Increasing and Non-Bounded Function I have made a solution for the following question and I'm wondering if it's correct. I think that something is missing here. Can you help me complete the solution?

Let $f$ be a function. The fuction $f$ is monotonic increasing and it is not bounded.
Prove that $$\lim_{x\to\infty} f(x)=\infty$$

Solution:
Assume in contradiction that $$\lim_{x\to\infty} f(x)=L$$
Therefore there is some $M\in \Bbb R$ such that $\left|f(x)-L\right|<\epsilon$.
Therefore, $L-\epsilon<f(x)<L+\epsilon$
We can now say the f(x) is bounded, which is a contradiction.
Something is missing here. I know that in sequenses, every convergent sequence is bounded. Is there a theorem for functions also?
Thanks,
Alan
 A: Let $M > 0$ be any given positive real number, since $f$ is unbounded, there is a natural number $N$ such that $f(N) > M$. Thus if $x > N \to f(x) > f(N) > M$. This shows $\displaystyle \lim_{x\to \infty} f(x) = +\infty$
A: I presume that this, holds whether the function is strictly increasing or not?
ie $F$ is only monotone increasing ( not' necessarily', strictly increasing) can one one say $x>N \rightarrow F(N)>F(x)$,? Or only $x>N \rightarrow F(x) \geq F(N)>M$?, 
One still has $F(x)>M$ even if $F(x)=F(N)$.; still holds here, of course.
I presume that one always find, an $N_1$ and, $M_1$, such that,  $M_1>\,[F(N)]\, > M$ where $F(N_1)>M_1$,  if $F$ is unbounded.
Thus as above one can keep on going and going finding larger and larger $N1$. If it $F$ were a constant, $F(x_i)=F(x_j)$for all $x_i>x_j$  on-wards,
It would contradict un-bounded-ness as one could not find, for all $N$ some F(x_k)>N. 
And  by monotone increasing alone;  
And, $F(N_1)>M_1>[F(N)=F(x)]\rightarrow F(N_1)>F(x)$ 
and $N_1>x$.
so that
$$\forall(x_2);[x_2>N_1] \rightarrow F(x_2)\geq F(N_1)>M_1>[F(x)=F(N)] $$
$$x_2>N_1>x>N$$ s.t $F(x_2) > F(N_1)>F(N) $; so $F(x_2)>$F(x)$ and one can keep on going and going. It would presumably be inconsistent with the function being a constant, its unbounded. So even its not strictly monotonic increasing I presume the above holds in the answer.  It clearly going to be increasing on the average, over and above again.
A: I presume it holds for an $M$ under the definition of unbounded-ness and so it would rule out a constant function by definition
For instance,   one such $N$ would be to set $N=F(x_i)$, as $N=F(x_i)$ and if the function is constant for all $x_k>x_j>x_i$one could not find $F(x_2)>N$ as $F(x_2)=F(x_i)=N$. So it would presumably have to diverge to infinity, if it unbounded, as if the function remained constant, one could always set the value of $N$ equal to said constant function value and that would contradict un-bounded-ness.
So long $\forall (x) F(x)=\infty$ 
Which probabily makes no sense, in any case  as a constant Bounded function ($F$ as defined is unbounded presumably by defintion, as the limit does not even exist).
