Do only many-one functions cross its horizontal asymptote? Informal definitions: many-one, rational function and asymptote
Many-one: For a function $f: A \to B$, If two o more elements of A, let's say, $a_{1}$ and $a_{2}$ have the same image in $B$, the function is referred to as a many-one function, also called many-to-one or non-injective since if a function is not one-one (also one-to-one), then it is many-one.
Rational function: Because functions can be expressed as a ratio between two polynomials, we can refer to them as rational functions, which is defined only when the denominator is non-zero. We also call them rational expressions
Asymptote: The line a cruve aproaches to as it heads towards infinity. Note that the distance between the line and the curve tends to zero as they head towards positive or negative infinity.
Finding the horizontal asymptote of a rational function
For the function  $h(x) = \frac{2x-3}{x+2}$, let $h(x)$ be equal to $y$, we then isolate x, thus we get $x = \frac{2y+3}{2-y}$, where $y \neq$ 2. Therefore, its horizontal asymptote is 2. By using this method, it may sometimes seem to result that you have two horizontal asymptotes, but graphically, there's only one.
A simpler way would be, since the degree of both polynomials are the same, we take the ratio of the leading coefficients as it is, in this case, 2.
In a similar manner, the horizontal asymptote of $g(x) = \frac{x-2}{x^{3} - 1}$ is zero since the degree of the denominator is greater than the numerator degree.
If you want to use limits, then we just evaluate the limit of the function as it goes to infinity.
How to tell if a function is many-one
Using the definition given for many-one, a function $f: A \to B$ is many-one if we consider two elements of $A$, for instance, $a_{1}$ and $a_{2}$ such that setting $f(a_{1})$ and $f(a_{2})$ equal to each other will yield $a_{1}$ $\neq$ $a_{2}$, otherwise the function is one-one. Thus, $h(x)$ is many-one, whereas $g(x)$ is not. You can also verify this through the horizontal line test, or do the computations.
When does a function cross the horizontal asymptote?
By definition, an asymptote is approached by the function as $x$ goes to infinity. If we were to consider finite values of x, not values of x that go to infinity, it may cross. By this I mean, $f(x)$ can be equal to the value the horizontal asymptote graphically represents for values of $x$ that do not approach infinity, or this may not happen as well. Because of this posibility, I ask this question, when does a function "cross" the horizontal asymptote?

Abramson, J. (2020, March 25). Rational functions. Mathematics LibreTexts https://math.libretexts.org/Courses/Borough_of_Manhattan_Community_College/MAT_206.5/04%3A_Polynomial_and_Rational_Functions/4.08%3A_Rational_Functions
As you can see, the curve crosses twice the horizontal asymptote. That is, zero is part of the range of the function, $f(x)$ can be equal to zero even though the horizontal asymptote is located at $y = 0$, which is expected. After all, the definition of an asymptote does only consider values of $x$ that go to infinity.
Note that, for other functions, it crosses once.

Rational functions: ratios of polynomials. (n.d.). Xaktly. Retrieved April 14, 2020, from https://xaktly.com/MathRationalFunctions.html
As I said before, it may also not cross. For instance, consider $h(x)$. The horizontal asymptote of $h(x)$ is located at $y = 2$, and since 2 is not part of the range of $h(x)$, it never crosses.

For the below, the function is $\sqrt{\frac{x-1}{x+1}}$, the horizontal asymptote is 1 (not 1 and -1)

Solution
For a rational function to cross its horizontal asymptote, the value the horizontal asymptote represents must be part of the range.
Thus, if the horizontal asymptote is equal to, let's say, $c$, then $c$ $\in$ $range$.
Consequently, $h(x)$ does not cross it horizontal asymptote because the range does not include 2.
What kind of rational functions cross the horizontal asymptote?
If you go back to the examples given, those who cross its horizontal asymptote are many-one functions and those who don't are one-one, which leads to my question.
Remark: I'm not considering rational trigonometric functions.
 A: Claim. If $f$ is a rational function with a horizontal asymptote at $y_0$, and there exists some $x_0$ with $f(x_0)=y_0$, then $f$ is many-to-one (i.e. not one-to-one).
Proof.  Note first that $f$ is continuous everywhere except at a finite number of points (the zeros of the denominator), so it is continuous on some neighborhood $(x_0 - \delta, x_0 + \delta)$ of $x_0$.  Pick two points $x_1, x_2$ with $x_0 - \delta < x_1 < x_0 < x_2 < x_0 + \delta$, and let $y_1 = f(x_1)$, $y_2 = f(x_2)$.  If $y_1 = y_0$ or $y_2 = y_0$ then we are done.  Otherwise, suppose without loss of generality that $y_1 < y_0$.  Now either $y_2 < y_0$ or $y_2 > y_0$.

*

*If $y_2 < y_0$, pick some number $t$ with $\max(y_1, y_2) < t < y_0$.  By the intermediate value theorem, there are points $s_1, s_2$ with $x_1 < s_1 < x_0 < s_2 < x_2$ with $f(s_1) = f(s_2) = t$ and we have that $f$ is many-to-one.


*If $y_2 > y_0$, then by the intermediate value theorem, $f$ attains every value in $[y_1, y_2]$ on the interval $[x_1, x_2]$.  Since $\lim_{x \to \pm \infty} f(x) = y_0$, there necessarily exists some $s$ outside $[x_1, x_2]$ with $f(s) \in [y_1, y_2]$.  So in this case as well, $f$ is many-to-one.
