Q. 86 - Calculus

Question

Use limit techniques to prove that a rational function $f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}$ will have,

1. a horizontal asymptote at $y = 0$, if the degree of p(x) is less than the degree of q(x);
2. a horizontal asymptote at $y=\frac{a_n}{b_m}$, where $a_n$ and $b_n$are the leading terms of p(x) and q(x), respectively, if p(x) and q(x) have the same degree;
3. no horizontal asymptote, if the degree of p(x) is greater than the degree of q(x).

Step-by-Step Solution

Answer

Part(a) It is proved that if the degree of p(x) is less than the degree of q(x) then the horizontal asymptote will be $y=0$.

Part(b) It is proved that a horizontal asymptote at $y=\frac{a_n}{b_m}$, where $a_n$ and $a_n$ are the leading terms of p(x) and q(x), respectively, if p(x) and q(x) have the same degree;

Part(c) It is proved that if the degree of p(x) is greater than the degree of q(x), then there is no horizontal asymptote.

Part(a) Step 1. Given Information

We are given a function,

$f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}$

Part(a) Step 1. Proving the statement.

Let $\mathrm{p}\left(\mathrm{x}\right)={\mathrm{a}}_0{x}^0+{\mathrm{a}}_1{x}^1+{\mathrm{a}}_2{x}^2+\dots \dots \dots +{\mathrm{a}}_m{x}^m$,

$$\begin{gathered} \mathrm{q}\left(\mathrm{x}\right)={\mathrm{b}}_0{x}^0+b_1{x}^1+b_2{x}^2+\dots \dots \dots +b_n{x}^n \\ y=f\left(x\right)=\frac{{\mathrm{a}}_0{x}^0+{\mathrm{a}}_1{x}^1+{\mathrm{a}}_2{x}^2+\dots \dots \dots +{\mathrm{a}}_m{x}^m}{{ \mathrm{b}}_0{x}^0+b_1{x}^1+b_2{x}^2+\dots \dots .+b_n{x}^n} \end{gathered}$$

The horizontal asymptotes of curves are horizontal lines that the graph of the function approaches to a number as $x\to \pm \infty$,

For $n>m$,

$$\begin{gathered} y=f\left(x\right) \\ =\frac{\frac{{\mathrm{a}}_0}{x^n}+\frac{a_1}{x^{n-1}}+\frac{a_2}{x^{n-2}}+\dots \dots \dots +\frac{a_m}{x^{n-2}}}{\frac{b_0}{x^n}+\frac{b_1}{x^{n-1}}+\frac{b_2}{x^{n-2}}+\dots \dots \dots +b_n} \\ \text{ as }x\to \pm \infty \\ y=f\left(x\right)\to \frac{0}{b_n} \\ y=f\left(x\right)\to 0 \end{gathered}$$

Hence, the given statement is proved.

Part(b) Step 1. Proving the statement.

For $n=m$,

$$\begin{gathered} y=f\left(x\right) \\ =\frac{\frac{a_0}{x^n}+\frac{a_1}{x^{n-1}}+\frac{a_2}{x^{n-2}}+\dots \dots \dots +a_m}{\frac{b_0}{x^n}+\frac{b_1}{x^{n-1}}+\frac{b_2}{x^{n-2}}+\dots \dots \dots +b_n} \\ \text{ as }x\to \pm \infty \\ y=f\left(x\right)\to \frac{{\mathrm{a}}_{\mathrm{m}}}{b_n} \end{gathered}$$

Hence, the given statement is proved.

Part(c) Step 1. Proving the statement.

For $n<m$,

$$\begin{gathered} y=f\left(x\right) \\ =\frac{\frac{a_0}{x^m}+\frac{a_1}{x^{m-1}}+\frac{a_2}{x^{m-2}}+\dots \dots \dots +a_m}{\frac{b_0}{x^m}+\frac{b_1}{x^{m-1}}+\frac{b_2}{x^{m-2}}+\dots \dots \dots +\frac{b_n}{x^{m-n}}} \\ \text{ as }x\to \pm \infty \\ y=f\left(x\right)\to \frac{a_m}{0} \\ y=f\left(x\right)\to \infty \end{gathered}$$

Hence, the given statement is proved.