On The Horizontal And Slanted Asymptotes Of Rational Functions

I am really confused about the horizontal and slant asymptotes of a function. My textbook says that given a rational function:

$$y=f(x)=\frac{a_{n}x^{n}+a_{n-1}x^{n-1}+\cdot\cdot\cdot+a_0}{b_{m}x^{m}+b_{m-1}x^{m-1}+\cdot\cdot\cdot+b_0}$$

The following properties are true. Can you please explain to me why the following properties are true? Do so not too rigorously, I'm still a beginner. And can you not use Calculus in the answers, since I haven't learn it yet? The properties are:

1. If $n < m$, then $y=0$ is a horizontal asymptote. I don't get why it would not be possible for 0 to be in the range of a rational function. Sure, a rational function like $\frac{1}{x}$ cannot produce 0. But how can this be true for all rational functions such that when $n < m$, it can't give an output of $0$? Please explain.

2. If $n>m$, there is no horizontal asymptote. Please explain to me why there would be no limit to what the function can produce.

3. If $n=m$, then $y=\frac{a_{n}}{b_{m}}$ is a horizontal asymptote. Again, I have no idea why this holds true for all rational functions. Can someone please explain?

4. If $n$ is more than $m$ only by $1$ degree, then there is a slant asymptote which can be determined by dividing the denominator into the numerator. Again, can someone please explain?

• "No calculus" implies no differentiation, which is fine; but in a comment, you also complained about the use of limits. Asymptotes essentially describe a "limit" behavior, so it seems difficult to avoid limits entirely. Perhaps you should explain what you think an asymptote is, and how you would be able to tell when a function's graph is asymptotic to a line. Commented Jun 30, 2018 at 18:18

The main result used head, in addition of Euclidean division of polynomials, t is that the limit of a rational function at $\infty$ is the limit of the ratio of the leading terms of the numerator and denominator, i.e. with your notations: $$\lim_{x\to\pm \infty} f(x)=\lim_{x\to\infty}\frac{a_nx^n} {b_mx^m}=\begin{cases}\displaystyle \lim_{x\to\infty}\frac{a_n} {b_mx^{m-n}}=0&\text{if }\:m>n,\\[0.5ex] \dfrac{a_n}{b_m}&\text{if }\:m=n,\\[0.5ex] \displaystyle \lim_{x\to\infty}\frac{a_n} {b_m}x^{n-m}=\infty&\text{if }\:m>n.\end{cases}$$

For the oblique asymptote, if $f(x)=\dfrac{g(x)}{h(x)}$, divide the numerator by the denominator: you obtain polynomials $q(x), \:r(x)$ such that $$g(x)=h(x)\,q(x)+r(x) \qquad r(x)=0~\text{ or }~\deg r <\deg h,$$ whence $$f(x)=q(x)+\frac{r(x)}{h(x)}, \quad\text{so }~f(x)-q(x)=\frac{r(x)}{h(x)}\to 0\quad\text{as }\: x\to\infty.$$

In this case we say the polynomial curve $y=q(x)$ is asymptote to the given rational curve $y=f(x)$.

In particular, if $\deg g=1+\deg h$, then $\deg r=1$ and the curve $y=r(x)$ is a straightline – which is called an oblique asymptote to the curve $y=f(x)$

• Thanks so much, but can you write an answer without calculus in it, because I haven't learnt it yet? Thanks. Commented Jun 30, 2018 at 14:01
• There's no calculus in it, just some algebra and the notion of limit. I wanted to insist on a more general notion of asymptote, which can be a curve, not only a straight line. Commented Jun 30, 2018 at 14:08
• Well I'm not very familiar with limit notation. Can you please simplify thanks? Also, just to check: x→∞ means x going up to infinity right? Commented Jun 30, 2018 at 14:09
• Yes (except it can be + \infinity or $-$ infinity, depending on what you mean. Commented Jun 30, 2018 at 14:13
• Okay thanks. Next thing though, can you explain limx→±∞ notation to me? I've seen this before but don't know what it means. Can you please explain? Or can you change the answer so that it doesn't need it. I know it is a lot to ask for but it will help. Commented Jun 30, 2018 at 14:16

(1) $0$ can be both an asymptotic value and in the range. They're unrelated. Consider $$f(x) = \frac{x}{x^2+1}$$ Certainly $f(0)=0$, but also $\lim\limits_{x\to\pm\infty}f(x)=0$.

(2) By performing polynomial division, such a function can be written in the form $$f(x) = p(x) + \frac{r(x)}{s(x)}$$ where $p$ is a polynomial of positive degree and $r$ and $s$ are polynomials of the form in (1). The polynomial part grows without bound as $x\to\pm\infty$, and $r/s$ tends to $0$ as in (1).

(3) Same argument as in (2), but here $p(x)$ is a polynomial of degree $0$ (constant).

(4) Again as in (1), but $p(x)$ is now a polynomial of degree $1$ (a nonhorizontal line).

• Thanks so much, but can you write an answer without calculus in it, because I haven't learnt it yet? Thanks. Commented Jun 30, 2018 at 13:41