Vtgyjfy

What is the cases that we use with it a horizontal asymptote ?
I mean is it always we use a horizontal asymptote we use $lim_{xto infty}$

limit as $fleft(xright)=:frac{x^2-1}{x^2+1}$ can not be $1$ which in this case it's obvious to use horizontal asymptote

But there is also some other functions like
$:fleft(xright)=left(frac{−3}{7}xright)$ which f(x) begin{bmatrix}mathrm{Solution:}:&:-infty :<fleft(xright)<infty \ :mathrm{Interval:Notation:}&:left(-infty :,:infty :right)end{bmatrix}

which not clear to use with it a horizontal asymptote !

what make me ask this Question is in my james book it defend a horizontal asymptote as :

The line $y=L$ is called a horizontal asymptote of the curve $y=f(x)$
if either

$lim :_{xto :infty ::}fleft(xright):=:L:$

or

$lim ::_{xto ::-infty :::}fleft(xright):=:L$

It's looks like a general definition for horizontal asymptote to use it with $lim_{xto infty}$

Horizontal asymptotes are found by $lim_limits{x to infty}f(x)$ and $lim_limits{x to -infty}f(x)$ given that they exist.

As an example, you have a rational function $frac{P(x)}{Q(x)}$ in which the two polynomials have the same degree:

$$f(x) = frac{a_1x^n+b_1x^{n-1}+c_1x^{n-2}+…+z_1}{a_2x^n+b_2x^{n-1}+c_2x^{n-2}+…+z_2}$$

As $x to infty$, $a_1x^n$ and $a_2x^n$ quickly outgrow the rest of the terms, leaving $frac{a_1x^n}{a_2x^n}$, so $lim_limits{x to pminfty}f(x) = frac{a_1}{a_2}$, which becomes the horizontal asymptote. This is usually shown through factoring and noting that $frac{c}{x} to 0$ as $x to infty$ for constant $c$:

$$f(x) = frac{left(a_1+frac{b_1}{x}+frac{c_1}{x^2}+…+frac{z_1}{x^n}right)}{x^nleft(a_2+frac{b_2}{x}+frac{c_2}{x^2}+…+frac{z_2}{x^n}right)} = frac{a_1+frac{b_1}{x}+frac{c_1}{x^2}+…+frac{z_1}{x^n}}{a_2+frac{b_2}{x}+frac{c_2}{x^2}+…+frac{z_2}{x^n}}$$

$$lim_{x to pm infty} f(x) = lim_{x to pm infty}frac{a_1color{blue}{+frac{b_1}{x}+frac{c_1}{x^2}+…+frac{z_1}{x^n}}}{a_2color{blue}{+frac{b_2}{x}+frac{c_2}{x^2}+…+frac{z_2}{x^n}}} = frac{a_1color{blue}{+0+0+…+0}}{a_2color{blue}{+0+0+…+0}} = frac{a_1}{a_2}$$

Hence, for $f(x) = frac{x^2+1}{x^2-1}$, the horizontal asymptote becomes $frac{1}{1} = 1$.

Your second example involves a linear function. A linear function has no asymptotes because its range is $y in mathbb{R}$ and $lim_limits{x to pm infty} f(x)$ is not a finite value. You’re looking for a value that $f(x)$ tends to as it approaches infinity or negative infinity, but that value may not always exist.

Sometimes, a function might have two horizontal asymptotes. An example of this is $f(x) = frac{3x+2}{sqrt{x^2-4}}$. To find the horizontal asymptotes, you use the normal process:

$$lim_{x to infty} f(x) = lim_{x to infty}frac{3x+2}{sqrt{x^2-4}} = lim_{x to infty}frac{xleft(3+frac{2}{x}right)}{sqrt{x^2left(1-frac{4}{x^2}right)}} = frac{x(3+0)}{sqrt{x^2(1+0)}} = frac{3x}{sqrt{x^2}}$$

Recalling $sqrt{x^2} = x$ for $x geq 0$, the first horizontal asymptote becomes $y = frac{3x}{x} = 3$. To find the second horizontal asymptote, you need $lim_limits{x to -infty} f(x)$:

$$lim_{x to -infty} f(x) = lim_{x to -infty}frac{3x+2}{sqrt{x^2-4}} = lim_{x to -infty}frac{xleft(3+frac{2}{x}right)}{sqrt{x^2left(1-frac{4}{x^2}right)}} = frac{3x}{sqrt{x^2}}$$

Recalling $sqrt{x^2} = -x$ for $x < 0$, the second horizontal asymptote becomes $y = frac{3x}{-x} = -3$. Here is the plot.

You can use the same process for other functions you are given.

There are functions like $f(x)=arctan(x)$ which have two horizontal asymptotes. Here both $y=pi/2$ [as $x to infty$] and $y=-pi/2$ [as $x to - infty$] are asymptotes. It can be seen by reflecting the graph of $tan(x)$ on $(-pi/2,pi/2)$ through the line $y=x.$

Another case for your question is $g(x)=e^x.$ It has $y=0$ as a horizontal asymptote, but only as $xto -infty.$ It gives an example where one cannot just consider $x to infty.$ [Here the limit as $x to infty$ is $infty,$ not a number $L$ as required in the book you mention.