When the values of the function \(f(x)\) approach the number \(L\) as \(x\) increases without a bound (meaning that the values of \(x\) can be chosen to be larger than any real number chosen),
\[\lim_{x\to+\infty}f(x)=L\]
Likewise,
\[\lim_{x\to-\infty}f(x)=M\]
when the functional values \(f(x)\) approach the number \(M\) as \(x\) decreases without a bound.
Horizontal Asymptotes
Geometrically, \(\lim_{x\to+\infty}f(x)=L^2\) means that the graph of \(f(x)\) approaches the horizontal line \(y=L\) at infinity, while \(\lim_{x\to-\infty}f(x)=M\) means that the graph of \(f(x)\) approaches the horizontal line \(y=m\) at negative infinity.
The lines \(y=L\) and \(y=M\) are called horizontal asymptotes.
Note that to find horizontal asymptotes, both the limit at \(+\infty\text{ and at }-\infty \) must be checked.
Reciprocal Power Rules
If \(A\) and \(p\) are constants with \(p>0\) and \(x^{p}\) is defined for all \(x\),then
$$\lim\limits_{x\to\infty}\frac{A}{x^{p}}=0\quad\text{and}\quad\lim\limits_{x\to-\infty}\frac{A}{x^{p}}=0\:.$$
Computation of Limits when \(\lim_{x\to\pm\infty}f(x)=\lim_{x\to\pm\infty}\frac{p(x)}{q(x)}\)
Step 1.
Divide each term in \(f(x)\) by the highest power of \(x^{p}\) that appears in the denominator. (Alternatively, divide by the highest power in the numerator, or by the highest power overall in \(f(x)\)).
Step 2.
Compute \(\lim_{x\to+\infty}f(x)\mathrm{~and~}\lim_{x\to-\infty}f(x)\) using algebraic properties of limits and the reciprocal power rules.
Example.
Evaluate \(\lim_{x\to\infty}\frac{3x^2-x-2}{5x^2+4x+1}\)
Solution.
The highest power in the denominator is \(x^{2}\). Divide the numerator and denominator by \(x^{2}\) to get:
\[\lim_{x\to\infty}\frac{3x^2-x-2}{5x^2+4x+1}=\lim_{x\to\infty}\frac{3-1/x-2/x^2}{5+4/x+1/x^2}=\frac{3-0-0}{5+0+0}=\frac{3}{5}\]
Thus the graph of \(f(x)=\frac{3x^2-x-2}{5x^2+4x+1}\) has a horizontal asymptote at \(y=\frac{3}{5}\). Note that the limit at \(-\infty\) gives the same answer.
Example.
Find \(\lim_{r\to-\infty}\frac{r^4-r^2+1}{r^5+r^3-r}.\)
Solution.
\[\lim_{r\to-\infty}\frac{r^4-r^2+1}{r^5+r^3-r}=\lim_{r\to-\infty}\frac{1/r-1/r^3+1/r^5}{1+1/r^2-1/r^4}=\frac{0-0+0}{1+0-0}=0\]
There is a horizontal asymptote at \(y=0\). Note that the limit at \(+\infty\) gives the same answer.
Example.
Find \(\lim_{x\to+\infty}\frac{x^2+x}{3-x}.\)
Solution.
\[\lim_{x\to+\infty}\frac{x^2+x}{3-x}=\lim_{x\to+\infty}\frac{x+1}{3/x-1}=-\infty \]
Thus, the graph of \(f(x)=\frac{x^2+x}{3-x}\) has no horizontal asymptote at \(+\infty\).
Note that to find the horizontal asymptotes, the limit at \(-\infty\) must be checked as well.