14.04 Limits at infinity

If the value of a function approaches some finite quantity as the domain variable becomes arbitrarily large, then that function value is loosely called a 'limit at infinity'. Such a function may have a graph that looks something like the following.

A function $f(x)$f(x) that approaches a finite limit as $x$x becomes large, is said to have a horizontal asymptote. 

A function may have an asymptote that is not horizontal. In such a case, the function does not approach a finite limit. That is, there is no limit at infinity. This can be seen by observing that such an asymptote is either vertical or it is a line with the form $y(x)=ax+b$y(x)=ax+b where a $\ne0$≠0, which is clearly unbounded as $x\rightarrow\infty$x→∞.

Certain rational functions have horizontal asymptotes. For this to occur, the denominator must be a polynomial of degree equal to or higher than the degree of the numerator polynomial.

To investigate the asymptotic behavior of rational functions, a useful fact to remember is the limit$\lim_{x\rightarrow\infty}\frac{1}{x}=0$limx→∞1x​=0. This is often used in combination with the limit laws discussed in our chapter on Evaluating limits and limit theorems.

In cases involving trigonometric functions, it can be helpful to recall these special limit theorems as well:

$\lim_{x\rightarrow0}\frac{\sin x}{x}=1$limx→0sinxx​=1 and $\lim_{x\rightarrow0}\frac{\cos x-1}{x}=0$limx→0cosx−1x​=0.

 

Worked examples

Question 1

Find the limit, if it exists, of the function $x\sin\frac{1}{x}$xsin1x​ as $x\rightarrow\infty$x→∞.

Think: As $x$x becomes arbitrarily large, the expression approaches the form $\infty\times0$∞×0 which is indeterminate. However, $x\sin\frac{1}{x}$xsin1x​ can be rewritten as $\frac{\sin\frac{1}{x}}{\frac{1}{x}}$sin1x​1x​​.

Do:

$\lim_{x\rightarrow\infty}x\sin\frac{1}{x}$limx→∞xsin1x​	$=$=	$\lim_{x\rightarrow\infty}\frac{\sin\frac{1}{x}}{\frac{1}{x}}$limx→∞sin1x​1x​​	Using the fact that $x=\frac{1}{\frac{1}{x}}$x=11x​​
	$=$=	$\lim_{a\rightarrow0}\frac{\sin a}{a}=1$lima→0sinaa​=1	Substituting $a=\frac{1}{x}$a=1x​ and as $x\to\infty$x→∞, $a\to0$a→0
	$=$=	$1$1	Using the fact that $\lim_{x\rightarrow0}\frac{\sin x}{x}=1$limx→0sinxx​=1 given above

 

Reflect:  Think about the graph of the function. Does our answer make sense?

 

Question 2

Investigate whether there are limits as $x\rightarrow-\infty$x→−∞ and $x\rightarrow\infty$x→∞ for the function $y(x)=\frac{x^3-2x}{2x^3+x^2-1}$y(x)=x3−2x2x3+x2−1​.

Think:  Does this function have any discontinuities?

Do:

$\lim_{x\rightarrow\infty}\frac{x^3-2x}{2x^3+x^2-1}$limx→∞x3−2x2x3+x2−1​	$=$=	$\lim_{x\rightarrow\infty}\frac{1-\frac{2}{x^2}}{2+\frac{1}{x}-\frac{1}{x^3}}$limx→∞1−2x2​2+1x​−1x3​​	Divide both the numerator and denominator by $x^3$x3
	$=$=	$\frac{\lim_{x\rightarrow\infty}1-\lim_{x\rightarrow\infty}\frac{2}{x^2}}{\lim_{x\rightarrow\infty}2+\lim_{x\rightarrow\infty}\frac{1}{x}-\lim_{x\rightarrow\infty}\frac{1}{x^3}}$limx→∞1−limx→∞2x2​limx→∞2+limx→∞1x​−limx→∞1x3​​	Using limit theorems
	$=$=	$\frac{1-0}{2+0-0}$1−02+0−0​	Using that $\lim_{x\rightarrow\infty}\frac{1}{x}=0$limx→∞1x​=0
	$=$=	$\frac{1}{2}$12​	Simplifying

 

It will not make any difference whether $x$x approaches positive or negative infinity as both $\lim_{x\rightarrow\infty}\frac{1}{x}=0$limx→∞1x​=0 and $\lim_{x\rightarrow-\infty}\frac{1}{x}=0$limx→−∞1x​=0. 

Therefore, the limit we seek is $\frac{1}{2}$12​.

Reflect: Can you think of another strategy that would involve rewriting $y(x)$y(x) in a different way?

 

Practice questions

Question 3

Question 4

question 5