A polynomial with integer coefficients maps integers to integers. A rational function (which is a quotient of two polynomials) tends to map integers to rational numbers. Other than the trivial case where the denominator is a factor of the numerator, are there any rational functions that map all integers to integers? What is known about them and how can they be found?
Rational functions from integers to integers
elementary-number-theoryrational-functions
Related Solutions
It always helps to divide/factor by $x$ raised to the highest power to see why the rules work the way they do. The important (yet obvious) thing to keep in mind is that for any constant $c$, $\dfrac{c}{x} \to 0$ as $x \to \infty$.
For the first case, you have the right idea. For instance, let $f(x) = \dfrac{x^2+5x+100}{x^3-5}$. You need to find $\lim_\limits{x \to \infty} f(x)$:
$$\lim_{x \to \infty}\frac{x^2+5x+100}{x^3-5} = \lim_{x \to \infty}\frac{x^2\left(\frac{5}{x}+\frac{100}{x^2}\right)}{x^3\left(1-\frac{5}{x^3}\right)} = \lim_{x \to \infty}\frac{1}{x} = 0$$
This is just another way of saying that the greatest powers will outgrow the other terms of the function as $x \to \infty$, and if the degree of the denominator is higher, clearly this means it tends to $0$. On the other hand, for the opposite case, the limit clearly tends to $\pm \infty$ (sign depends on the sign of the leading coefficients of the numerator and denominator).
For the second case, you can once again show this through factoring. For instance, say you have a function in the form $f(x) = \dfrac{ax^2+bx+c}{dx^2+ex+f}$. You once again need to find $\lim_\limits{x \to \infty} f(x)$:
$$\lim_{x \to \infty} \frac{ax^2+bx+c}{dx^2+ex+f} = \lim_{x \to \infty} \frac{x^2\left(a+\frac{b}{x}+\frac{c}{x^2}\right)}{x^2\left(d+\frac{e}{x}+\frac{f}{x^2}\right)} = \frac{a}{d}$$
You can think of this as the $ax^2$ and $dx^2$ terms eventually outgrowing all the other terms as $x \to \infty$, so all that “remains” is $\dfrac{ax^2}{dx^2} = \dfrac{a}{d}$.
Can you use similar arguments to show the third case (such as when the numerator is a cubic and the denominator is a quadratic)? As a start, you can notice that the terms with lower powers will eventually vanish in both the numerator and denominator, and you can try to justify this with a limit.
Addition: You got the idea. Using the limit, it becomes apparent that due to the numerator begin a single power higher, all that will “remain” as $x \to \infty$ is $mx$. As you saw, however, the limit gives only the slope $m$ and not the $y$-intercept ($b$) of the slanted asymptote, which is in the form $y = mx+b$. For this purpose, you would have to calculate another limit as well:
$$\lim_{x \to \infty} f(x)-mx = b$$
This gives the difference between $f(x)$ and $mx$ as $x \to \infty$, which will give $b$:
$$\implies \lim_{x \to \infty} \frac{3x^3+6x^2+4x+2}{x^2+5x+2}-3x = \lim_{x \to \infty} \frac{3x^3+6x^2+4x+2-3x^3-15x^2-6x}{x^2+5x+2}$$
Simplifying and using case $(2)$, you get
$$\lim_{x \to \infty} \frac{-9x^2-2x+2}{x^2+5x+2} = -9$$
Putting it all together, you get $y = 3x-9$, which is the same result obtained via polynomial division.
Because of their similarity to rational numbers; i.e.---
A function $f(x)$ is called a rational function provided that
$$f(x) = \frac{P(x)}{Q(x)},$$
where $P(x)$ and $Q(x)$ are polynomials and $Q(x)$ is not the zero polynomial; just like a real number $n$ is called a rational number, provided that
$$ n = \frac{p}{q},$$
where $p, q$ are integers and $q \neq 0$.
Best Answer
Let $r(z) = \frac{p(z)}{q(z)}$ such that $r:\mathbb{Z}\rightarrow\mathbb{Z}$ and $p,q$ polynomials. Then $q(z)|p(z)$ for all $z \in \mathbb{Z}$ and $\deg p \geq \deg q$ (check what happens at $\infty$).
You can check here, for example, how the division between polynomials over a commutative ring (like $\mathbb{Z}$) works.
So if $q$ is monic, then you can divide the polynomials, so $p(z)=f(z)q(z)+g(z)$ with $f,g$ polynomials and $\deg g < \deg q$. Then if $g \not \equiv 0$, $r(z)=f(z)+\frac{g(z)}{q(z)}$, so $\frac{g(z)}{q(z)}$ is also an "integer" rational function, which is absurd. So $g\equiv0$ and $r$ is a integer polynomial.
If $q$ is non-monic, you can apply the theorem of the linked answer: call $q_0$ the leading coefficient of $q$, then $q_0^kp(z) = f(z)q(z) + g(z)$. Consider the integer rational function $q_0^kr(z) = \frac{q_0^kp(z)}{q(z)} = f(z)+ \frac{g(z)}{q(z)}$. As above, this implies that $g(z) \equiv 0$. So $r(z)=\frac{p(z)}{q(z)}=\frac{q_0^kp(z)}{q_0^kq(z)}=\frac{f(z)q(z)}{q_0^kq(z)}=\frac{f(z)}{q_0^k}$. Then again $r$ is a polynomial, possibly with rational coefficients ($f$ is a integer polynomial, but might not be divisible by $q_0^k$).