Is there a theorem in real analysis that a strictly increasing function must cross a certain value or have a horizontal asymptote smaller than that? Is there a theorem in real analysis that a strictly increasing function must cross a certain value or have a horizaontal asymptote smaller than that value? 
EDIT: Let's say the function is continuous, and crossing means that the for some x, f(x) = y, where y is the aformentioned value.
 A: Do you mean to ask about the statement that:

For any value $y$ one of the following holds: Either $\lim_{x\rightarrow\infty}f(x)\leq y$ or there exists some triple of $a$, $b$, and $c$ such that $f(a)<f(b)=y<f(c)$? 

If so, it's not quite true; we could consider the example of the function
$$f(x)=\begin{cases}\frac{x}{x+1}+1&&\text{if }x>0\\\frac{-x}{x-1}-1 &&\text{otherwise}\end{cases}$$
which, over the interval $(-\infty,0]$ has values in $(-2,-1]$ and over $(0,\infty)$ has values in $(1,2)$, but is strictly increasing everywhere. It never "crosses" any values in $(-1,1]$, since it never obtains such values. Even if you count the discontinuity as a crossing of all those values (i.e. since it was less, and then it was more, we say it crossed, even though it never really did), it still happens that for values like $y=-3$, the function never crosses it, being everywhere greater than $-3$, but it certainly has no asymptote below $-3$ either.
What could be said is that, either $f(x)$ is always less than $y$ (are hence has an asymptote less than or equal to $y$), or $f(x)$ is somewhere greater than $y$, but this is almost tautological.
