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

\begin{equation} 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} \end{equation}

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?

Thank you so much in advance. Again, please try not to use calculus in your answers.

  • $\begingroup$ "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. $\endgroup$ – David K Jun 30 '18 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 $y=f(x)$

  • $\begingroup$ Thanks so much, but can you write an answer without calculus in it, because I haven't learnt it yet? Thanks. $\endgroup$ – Ethan Chan Jun 30 '18 at 14:01
  • $\begingroup$ 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. $\endgroup$ – Bernard Jun 30 '18 at 14:08
  • $\begingroup$ 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? $\endgroup$ – Ethan Chan Jun 30 '18 at 14:09
  • $\begingroup$ Yes (except it can be + \infinity or $-$ infinity, depending on what you mean. $\endgroup$ – Bernard Jun 30 '18 at 14:13
  • $\begingroup$ 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. $\endgroup$ – Ethan Chan Jun 30 '18 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).

  • $\begingroup$ Thanks so much, but can you write an answer without calculus in it, because I haven't learnt it yet? Thanks. $\endgroup$ – Ethan Chan Jun 30 '18 at 13:41

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