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 behaviour 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 another chapter.

In cases involving trigonometric functions, it can be helpful to recall the limits $\lim_{x\rightarrow0}\frac{\sin x}{x}=1$limx→0sinxx​=1 and $\lim_{x\rightarrow0}\frac{\cos x-1}{x}=0$limx→0cosx−1x​=0.

 

Example 1

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

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​​.

We know that $\lim_{y\rightarrow0}\frac{\sin y}{y}=1$limy→0sinyy​=1 and $\lim_{x\rightarrow\infty}\frac{1}{x}=0$limx→∞1x​=0. Now, if we put $y=\frac{1}{x}$y=1x​, we have $\lim_{x\rightarrow\infty}\frac{\sin\frac{1}{x}}{\frac{1}{x}}=\lim_{y\rightarrow0}\frac{\sin y}{y}=1$limx→∞sin1x​1x​​=limy→0sinyy​=1.

 

Example 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​.

We rewrite the function as $\frac{1-\frac{2}{x^2}}{2+\frac{1}{x}-\frac{1}{x^3}}$1−2x2​2+1x​−1x3​​ by dividing the numerator and denominator by $x^3$x3. Now, because $\lim_{x\rightarrow\infty}\frac{1}{x}=0$limx→∞1x​=0 it must be true (by a limit theorem) that $\lim_{x\rightarrow\infty}\frac{1}{x^2}=0$limx→∞1x2​=0 and $\lim_{x\rightarrow\infty}\frac{1}{x^3}=0$limx→∞1x3​=0. It will not make any difference whether $x$x approaches positive or negative infinity.

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

Example 3

Example 4

Example 5