Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 3

Limits

Limits at Infinity: Horizontal Asymptotes

We know how to apply theorems to evaluate limits at a finite number, but now we must learn how to evaluate limits at infinity. We use these limits to find the horizontal asymptotes of functions. We begin by learning the definition of a horizontal asymptote from the graph of the function.

The graph of a function with a horizontal asymptote $y=L$ in the positive direction may look like that shown in Figure 3.26, below (see also Figure 2 on page 85 of the textbook).

Figure 3.26. Function with horizontal asymptote $y=L$ in the positive direction

The graph of a function with a horizontal asymptote $y=L$ in the negative direction may look like in Figure 3.27, below (see also Figure 3 on page 85 of the textbook).

Figure 3.27. Function with horizontal asymptote $y=L$ in the negative direction

What makes the line $y=L$ an asymptote is the fact that the values of the function $f\left(x\right)$ are close to $L$, and the larger and positive or smaller and negative $x$ is, the closer $f\left(x\right)$ is to $L\text{.}$ To indicate that $x$ is very large and positive—that is, that $x$ increases without bound—we write $x\to \infty \text{.}$ To indicate that $x$ is very small and negative—that is, that $x$ decreases without bound—we write $x\to -\infty \text{.}$

It is a misconception to say that a function has a horizontal asymptote when the function gets close to the horizontal line, but never touches it. As you see in Figure 2 on page 85 of the textbook, the function oscillates around the horizontal line, and the values of the function get close to or are equal to $L\text{.}$

Remember: The symbols $\infty$ and $-\infty$ do not represent numbers, and we must interpret the notation $x\to \infty$ and $x\to -\infty$ correctly. Even when we say that “$x$ approaches infinity,” we mean that $x$ increases without bound, no more and no less.

PowerPoint 1: For a large x, the values of the function f(x) tend to M. Hence y = M is a horizontal asymptote.

PowerPoint 2: For a negative small x, the values of the function f(x) tend to M. Hence y = M is a horizontal asymptote.

Example 3.63. From Figure 3.19, we see that $y=0$ is a horizontal asymptote in both directions of the function ${\displaystyle f\left(x\right)={\displaystyle \frac{1}{x}}\text{.}}$ That is

${\displaystyle \underset{x\to \infty }{\lim }{\displaystyle \frac{1}{x}}=0\text{and}\underset{x\to -\infty }{\lim }{\displaystyle \frac{1}{x}}=0\text{.}}$

Example 3.64. From the graph of the function $f$ shown in Figure 3.28, below, we see that the vertical lines $x=M$ and $x=0$ are vertical asymptotes, and the horizontal line $y=L$ is a horizontal asymptote.

In terms of limits,

$\underset{x\to M^{-}}{\lim }f\left(x\right)=\infty \text{,}\underset{x\to 0^{+}}{\lim }f\left(x\right)=\infty \text{,}\underset{x\to \infty }{\lim }f\left(x\right)=L\text{,}\underset{x\to -\infty }{\lim }f\left(x\right)=\infty \text{.}$

Figure 3.28. Function with two vertical asymptotes and one horizontal asymptote

Observe that, by Definition 3.27, for us to take a limit at infinity, the function must be defined on an interval $\left(c,\infty \right)$, for some $c$; and for us to evaluate limits at negative infinity, the function must be defined on an interval $\left(-\infty ,d\right)$ for some number $d\text{.}$ For instance, we cannot take the limit at infinity or negative infinity of the tangent, cotangent, secant or cosecant functions. We also see that the sine and cosine functions oscillate between $-1$ and $1$ as $x$ increases (positively) or decreases (negatively). Therefore, the limits

$\underset{x\to \infty }{\lim }\sin x\text{,}\underset{x\to -\infty }{\lim }\sin x\text{,}\underset{x\to \infty }{\lim }\cos x\text{,}\text{and}\underset{x\to -\infty }{\lim }\cos x$

do not exist.

Some of the definitions and theorems we apply to evaluate finite limits also hold for infinite limits. More precisely, the following definition and results are valid if we replace $x\to a$ (and $x\to b$ in Theorem 3.20) by either $x\to \infty$ or $x\to -\infty \text{.}$

1. Definition 3.9
2. Proposition 3.16
3. Theorem 3.17—parts (e) and (f) (limit of the reciprocal of a function)
4. Theorem 3.18
5. Theorem 3.20 (composition of functions—infinity limits)
6. Theorem 3.21 (sum and product of infinity limits)
7. Theorem 3.22 (Squeeze Theorem)
8. Theorem 3.23 (Laws of Limits)
9. Theorem 3.24
10. Corollary 3.25
11. Corollary 3.26

Example 3.65. (Theorem 4 on page 87 of the textbook).

The function

${\displaystyle g\left(x\right)={\displaystyle \frac{1}{x}}}$

is continuous on the intervals $\left(-\infty ,0\right)$ and $\left(0,\infty \right)\text{.}$

If $r$ is a positive rational number, then the function $f\left(x\right)=x^r$ is continuous at $0$ from the right, and if $u=1∕x$, then $u\to 0^{+}$ as $x\to \infty \text{.}$ [Why?]

By Corollary 3.26,

${\displaystyle \underset{x\to \infty }{\lim }{\displaystyle \frac{1}{x^r}}=\underset{u\to 0^{+}}{\lim }{u}^r=0\text{.}}$

Similarly, if $r$ is a positive rational number such that $f\left(x\right)=x^r$ is defined on an interval $\left(-\infty ,c\right)$ for some $c\text{,}$ then

${\displaystyle \underset{x\to -\infty }{\lim }{\displaystyle \frac{1}{x^r}}=\underset{u\to 0^{-}}{\lim }{u}^r=0\text{.}}$

A particular case of Example 3.65 occurs when $n$ is any positive integer. Then

${\displaystyle f\left(x\right)={\displaystyle \frac{1}{x^n}}}$

is defined on the intervals $\left(-\infty ,0\right)$ and $\left(0,\infty \right)$; and therefore,

${\displaystyle \underset{x\to \infty }{\lim }{\displaystyle \frac{1}{x^n}}=0\text{and}\underset{x\to -\infty }{\lim }{\displaystyle \frac{1}{x^n}}=0\text{.}}$

Example 3.66. Example 3.65, with $r=1/3\text{,}$ says that

${\displaystyle \underset{x\to -\infty }{\lim }{\displaystyle \frac{1}{\sqrt[3]{x}}}=0\text{.}}$

By Corollary 3.25,

${\displaystyle \underset{x\to -\infty }{\lim }\sin \left({\displaystyle \frac{1}{\sqrt[3]{x}}}\right)=0\text{.}}$

To evaluate the limits at infinity of rational functions, we must remember that a polynomial function of degree $n$ with leading coefficient $a_n$ is of the form

$P\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+....+a_{1}x+a_0\text{.}$

Example 3.67. Let us identify the degree and the leading coefficient of a few polynomial functions.

1. $Q\left(x\right)=3-x+6x^4$, degree $4$, leading coefficient $6\text{.}$
2. $P\left(x\right)=-x^3+6x^2+x$, degree $3$, leading coefficient $-1\text{.}$
3. $P\left(x\right)=6$, degree $0$, leading coefficient $6\text{.}$

In the next three examples, we consider the strategy used to evaluate infinite limits of rational functions.

Example 3.68. For the limit

${\displaystyle \underset{x\to \infty }{\lim }{\displaystyle \frac{3-x-x^2}{2x^2+x}}\text{,}}$

we identify the largest power of the polynomials in the numerator and denominator—in this case, $2\text{.}$ We divide the numerator and the denominator by $x^2\text{,}$ and we find that

${\displaystyle {\displaystyle \frac{3-x-x^2}{2x^2+x}}={\displaystyle \frac{{\displaystyle \frac{3}{x^2}}-{\displaystyle \frac{1}{x}}-1}{2+{\displaystyle \frac{1}{x}}}}\text{.}}$

We apply Proposition 3.16, Theorem 3.23 and Example 3.65, above, to conclude that

${\displaystyle \begin{array}{ll}\underset{x\to \infty }{\lim }{\displaystyle \frac{3-x-x^2}{2x^2+x}}\hfill & =\underset{x\to \infty }{\lim }{\displaystyle \frac{{\displaystyle \frac{3}{x^2}}-{\displaystyle \frac{1}{x}}-1}{2+{\displaystyle \frac{1}{x}}}}\hfill \\ \hfill & ={\displaystyle \frac{\underset{x\to \infty }{\lim }{\displaystyle \frac{3}{x^2}}-{\displaystyle \frac{1}{x}}-1}{\underset{x\to \infty }{\lim }2+{\displaystyle \frac{1}{x}}}}=-{\displaystyle \frac{1}{2}}\text{.}\hfill \end{array}}$

The line ${\displaystyle y=-{\displaystyle \frac{1}{2}}}$ is a horizontal asymptote.

Observe that when the degree of the numerator and denominator are equal, the limit is $-1$ (the leading coefficient of the numerator polynomial) over $2$ (the leading coefficient of the denominator polynomial).

Example 3.69. In this example, we proceed as in Example 3.68, above.

In the rational function

${\displaystyle {\displaystyle \frac{3-x}{2x^2+x}}\text{,}}$

the biggest power is $2\text{.}$ Dividing both the numerator and the denominator by $x^2$, we obtain

${\displaystyle {\displaystyle \frac{3-x}{2x^2+x}}={\displaystyle \frac{{\displaystyle \frac{3}{x^2}}-{\displaystyle \frac{1}{x}}}{2+{\displaystyle \frac{1}{x}}}}\text{.}}$

Applying Proposition 3.16, Theorem 3.23 and Example 3.65, we find that

${\displaystyle \begin{array}{ll}\underset{x\to -\infty }{\lim }{\displaystyle \frac{3-x}{2x^2+x}}\hfill & =\underset{x\to -\infty }{\lim }{\displaystyle \frac{{\displaystyle \frac{3}{x^2}}-{\displaystyle \frac{1}{x}}}{2+{\displaystyle \frac{1}{x}}}}\hfill \\ \hfill & ={\displaystyle \frac{\underset{x\to -\infty }{\lim }{\displaystyle \frac{3}{x^2}}-{\displaystyle \frac{1}{x}}}{\underset{x\to -\infty }{\lim }2+{\displaystyle \frac{1}{x}}}}={\displaystyle \frac{0}{2}}=0.\hfill \end{array}}$

The line $y=0$ is a horizontal asymptote.

In this case, the degree of the numerator is less than the degree of the denominator.

Example 3.70. In the limit

${\displaystyle \underset{x\to -\infty }{\lim }{\displaystyle \frac{3-x^3}{2x^2+x}}\text{,}}$

the largest degree is $3\text{.}$

Dividing both the numerator and the denominator by $x^3$, we obtain

${\displaystyle {\displaystyle \frac{3-x^3}{2x^2+x}}={\displaystyle \frac{{\displaystyle \frac{3}{x^3}}-1}{{\displaystyle \frac{2}{x}}+{\displaystyle \frac{1}{x^2}}}}=\left({\displaystyle \frac{3}{x^3}}-1\right)\left({\displaystyle \frac{1}{{\displaystyle \frac{2}{x}}+{\displaystyle \frac{1}{x^2}}}}\right)\text{.}}$

We know that

1. ${\displaystyle \underset{x\to -\infty }{\lim }{\displaystyle \frac{3}{x^3}}-1=-1<0}$ by Example 3.65 and Theorem 3.23.
2. ${\displaystyle \underset{x\to -\infty }{\lim }{\displaystyle \frac{2}{x}}+{\displaystyle \frac{1}{x^2}}=0}$ by Example 3.65 and Theorem 3.23.
3. ${\displaystyle \frac{2}{x}}+{\displaystyle \frac{1}{x^2}}<0$ for $x\to -\infty (x<0)$.
4. ${\displaystyle \underset{x\to -\infty }{\lim }{\displaystyle \frac{1}{{\displaystyle {\displaystyle \frac{2}{x}}+{\displaystyle \frac{1}{x^2}}}}}=-\infty }$ by (a) and (b), above, and Theorem 3.17(f).

Applying Theorem 3.18, we obtain

${\displaystyle \underset{x\to -\infty }{\lim }{\displaystyle \frac{3-x^3}{2x^2+x}}=\underset{x\to -\infty }{\lim }\left({\displaystyle \frac{3}{x^3}}-1\right)\left({\displaystyle \frac{1}{{\displaystyle \frac{2}{x}}+{\displaystyle \frac{1}{x^2}}}}\right)=\infty \text{.}}$

Observe that the rational function

${\displaystyle R\left(x\right)={\displaystyle \frac{3-x^3}{2x^2+x}}>0\text{,}}$

for $x\to -\infty$, $\left(x<0\right)\text{.}$ To see this, take a very small negative value of $x$, say $x=-1000$, and check that $R\left(-1000\right)>0\text{.}$

In this case, the degree of the numerator is greater than the degree of the denominator.

The strategy we used in the last three examples should convince us that the degree of the numerator and denominator determine the limits at infinity of rational functions. So, we have the following result.

Theorem 3.29. If ${\displaystyle R\left(x\right)={\displaystyle \frac{p\left(x\right)}{q\left(x\right)}}}$ is a rational function, then

1. ${\displaystyle \underset{x\to \pm \infty }{\lim }{\displaystyle \frac{p(x)}{q(x)}}={\displaystyle \frac{\text{leading}\text{coefficient}\text{of}p(x)}{\text{leading}\text{coefficient}\text{of}q(x)}}}$ if degree of $p(x)=$ degree of $q(x)$.
2. ${\displaystyle \underset{x\to \pm \infty }{\lim }{\displaystyle \frac{p(x)}{q(x)}}=0}$ if degree of $p(x)<$ degree of $q\left(x\right)\text{.}$
3. ${\displaystyle \underset{x\to \infty }{\lim }{\displaystyle \frac{p(x)}{q(x)}}=\infty (-\infty )}$ if degree of $p\left(x\right)>$ degree of $q\left(x\right)$ and ${\displaystyle R\left(x\right)={\displaystyle \frac{p\left(x\right)}{q\left(x\right)}}>0\left(<0\right)}$ for $x\to \infty \text{.}$
4. ${\displaystyle \underset{x\to -\infty }{\lim }{\displaystyle \frac{p(x)}{q(x)}}=\infty (-\infty )}$ if degree of $p\left(x\right)>$ degree of $q\left(x\right)$ and ${\displaystyle R\left(x\right)={\displaystyle \frac{p\left(x\right)}{q\left(x\right)}}>0\left(>0\right)}$ for $x\to -\infty \text{.}$

Therefore, we see that a rational function has a horizontal asymptote only when the degree of the numerator is less than or equal to the degree of the denominator.

Example 3.71. From Theorem 3.29, we have

${\displaystyle \underset{x\to \infty }{\lim }{\displaystyle \frac{\pi x}{2x+1}}={\displaystyle \frac{\pi }{2}}}$

if ${\displaystyle t={\displaystyle \frac{\pi x}{2x+1}}}$, then ${\displaystyle t\to {\displaystyle \frac{\pi }{2}}}$ as $x\to \infty \text{.}$

Hence, by Corollary 3.26 and Theorem 3.14,

${\displaystyle \underset{x\to \infty }{\lim }{\left({\displaystyle \frac{\pi x}{2x+1}}\right)}^2=\underset{t\to {\displaystyle \frac{\pi }{2}}}{\lim }{t}^2={\displaystyle \frac{{\pi }^2}{4}}\text{.}}$

Example 3.72. From the definition of absolute value, we have

$\sqrt{x^2}=x\text{for}x\ge 0\text{,}\text{and}\sqrt{x^2}=-x\text{for}x<0\text{.}$

Hence,

${\displaystyle \sqrt{3x^2+4x}=\sqrt{x^2\left(3+{\displaystyle \frac{4}{x}}\right)}=x\sqrt{3+4x}\text{for}x\to \infty \text{,}}$

and

${\displaystyle \sqrt{3x^2+4x}=\sqrt{x^2\left(3+{\displaystyle \frac{4}{x}}\right)}=-x\sqrt{3+4x}\text{for}x\to -\infty \text{.}}$

By Proposition 3.16, Corollary 3.26, Theorem 3.23 and Example 3.65:

${\displaystyle \begin{array}{ll}\underset{x\to \infty }{\lim }{\displaystyle \frac{\sqrt{3x^2+4x}}{x+7}}\hfill & =\underset{x\to \infty }{\lim }{\displaystyle \frac{x\sqrt{3+{\displaystyle \frac{4}{x}}}}{x\left(1+{\displaystyle \frac{7}{x}}\right)}}=\underset{x\to \infty }{\lim }{\displaystyle \frac{\sqrt{3+{\displaystyle \frac{4}{x}}}}{\left(1+{\displaystyle \frac{7}{x}}\right)}}\hfill \\ \hfill & ={\displaystyle \frac{\sqrt{\underset{x\to \infty }{\lim }3+{\displaystyle \frac{4}{x}}}}{\underset{x\to \infty }{\lim }\left(1+{\displaystyle \frac{7}{x}}\right)}}=\sqrt{3}\hfill \end{array}}$

and

${\displaystyle \begin{array}{ll}\underset{x\to -\infty }{\lim }{\displaystyle \frac{\sqrt{3x^2+4x}}{x+7}}\hfill & =\underset{x\to -\infty }{\lim }{\displaystyle \frac{-x\sqrt{3+{\displaystyle \frac{4}{x}}}}{x\left(1+{\displaystyle \frac{7}{x}}\right)}}=\underset{x\to -\infty }{\lim }{\displaystyle \frac{-\sqrt{3+{\displaystyle \frac{4}{x}}}}{\left(1+{\displaystyle \frac{7}{x}}\right)}}\hfill \\ \hfill & ={\displaystyle \frac{-\sqrt{\underset{x\to -\infty }{\lim }3+{\displaystyle \frac{4}{x}}}}{\underset{x\to -\infty }{\lim }\left(1+{\displaystyle \frac{7}{x}}\right)}}=-\sqrt{3}\hfill \end{array}}$

Example 3.73. For $x\to \infty$

\begin{align*} \dfrac{3x^3 - 3x + 1}{\sqrt{2x^2 + 3}} & = \dfrac{x^3\left(3 - \displaystyle{\frac{3}{x^2}} + \displaystyle{\frac{1}{x^3}}\right)}{\sqrt{x^2\left(2 + \displaystyle{\frac{3}{x^2}}\right)}} \\ & = \frac{x^3\left(3 - \displaystyle{\frac{3}{x^2}} + \displaystyle{\frac{1}{x^3}}\right)}{x\sqrt{2 + \displaystyle{\frac{3}{x^2}}}} \\ & = x^2\left(\frac{3 - \displaystyle{\frac{3}{x^2}} + \displaystyle{\frac{1}{x^3}}}{\sqrt{2 + \displaystyle{\frac{3}{x^2}}}}\right) \! {\text{.}} \end{align*}

1. ${\displaystyle \underset{x\to \infty }{\lim }{x}^2=\infty }$ by Theorem 3.21(a).
2. $\displaystyle{\lim_{x \to \infty} \frac{3 - \displaystyle{\frac{3}{x^2}} + \displaystyle{\frac{1}{x^3}}}{\sqrt{2 + \displaystyle{\frac{3}{x^2}}}} = \frac{3}{\sqrt{2}} \gt 0}$ by Corollary 3.26, Theorem 3.23 and Example 3.65.

From points (a) and (b), above, and Theorem 3.21, we have

${\displaystyle \underset{x\to \infty }{\lim }{x}^2\left({\displaystyle \frac{3-{\displaystyle \frac{3}{x^2}}+{\displaystyle \frac{1}{x^3}}}{\sqrt{2+{\displaystyle \frac{3}{x^2}}}}}\right)=\infty }$

and by Proposition 3.16,

${\displaystyle \underset{x\to \infty }{\lim }{\displaystyle \frac{3x^3-3x+1}{\sqrt{2x^2+3}}}=\underset{x\to \infty }{\lim }{x}^2\left({\displaystyle \frac{3-{\displaystyle \frac{3}{x^2}}+{\displaystyle \frac{1}{x^3}}}{\sqrt{2+{\displaystyle \frac{3}{x^2}}}}}\right)=\infty \text{.}}$

Note: In the last two examples, we factored the $\color{#384877}{x}$ with the biggest power, and simplified until we could apply one or several of the results listed above.

Example 3.74. For the limit

$\underset{x\to -\infty }{\lim }{x}^2-\sqrt{2x^4+x^2}\text{,}$

we use a different strategy. Here, we multiply by the conjugate, and then we proceed as in the previous examples.

${\displaystyle \begin{array}{ll}\left(x^2-\sqrt{2x^4+x^2}\right)\left({\displaystyle \frac{x^2+\sqrt{2x^4+x^2}}{x^2+\sqrt{2x^4+x^2}}}\right)\hfill & ={\displaystyle \frac{x^4-\left(2x^4+x^2\right)}{x^2+\sqrt{2x^4+x^2}}}\hfill \\ \hfill & ={\displaystyle \frac{-x^4-x^2}{x^2+\sqrt{2x^4+x^2}}}\hfill \end{array}}$

Factoring,

${\displaystyle \begin{array}{ll}{\displaystyle \frac{-x^4-x^2}{x^2+\sqrt{2x^4+x^2}}}\hfill & ={\displaystyle \frac{-x^4\left(1+{\displaystyle \frac{1}{x^2}}\right)}{x^2\left(1+\sqrt{2+{\displaystyle \frac{1}{x^2}}}\right)}}\hfill \\ \hfill & =-{\displaystyle \frac{x^4\left(1+{\displaystyle \frac{1}{x^2}}\right)}{x^2\left(1+\sqrt{2+{\displaystyle \frac{1}{x^2}}}\right)}}=-x^2\left({\displaystyle \frac{1+{\displaystyle \frac{1}{x^2}}}{1+\sqrt{2+{\displaystyle \frac{1}{x^2}}}}}\right)\hfill \end{array}}$

1. $\underset{x\to -\infty }{\lim }{x}^2=\infty$, so, $\underset{x\to -\infty }{\lim }-x^2=-\infty$ by Theorem 3.21(a).
2. $\underset{x\to -\infty }{\lim }{\displaystyle \frac{1+{\displaystyle \frac{1}{x^2}}}{1+\sqrt{2+{\displaystyle \frac{1}{x^2}}}}}={\displaystyle \frac{1}{1+\sqrt{2}}}>0$ by Corollary 3.26, Theorem 3.23 and Example 3.65.

From points (a) and (b), above, and Theorem 3.21, we have

${\displaystyle \underset{x\to -\infty }{\lim }-x^2\left({\displaystyle \frac{1+{\displaystyle \frac{1}{x^2}}}{1+\sqrt{2+{\displaystyle \frac{1}{x^2}}}}}\right)=-\infty }$

and by Proposition 3.16,

${\displaystyle \underset{x\to -\infty }{\lim }{x}^2-\sqrt{2x^4+x^2}=\underset{x\to -\infty }{\lim }-x^2\left({\displaystyle \frac{1+{\displaystyle \frac{1}{x^2}}}{1+\sqrt{2+{\displaystyle \frac{1}{x^2}}}}}\right)=-\infty \text{.}}$

Example 3.75. To evaluate the limit

${\displaystyle \underset{x\to \infty }{\lim }{\displaystyle \frac{\cos x}{x^2}}\text{,}}$

we use the fact that, for any nonzero $x$, $-1\le \cos x\le 1\text{.}$ Since $x^2>0$, we multiply by ${\displaystyle \frac{1}{x^2}}\text{,}$ and we get

${\displaystyle -{\displaystyle \frac{1}{x^2}}\le {\displaystyle \frac{\cos x}{x^2}}\le {\displaystyle \frac{1}{x^2}}\text{.}}$

By Example 3.65,

${\displaystyle \underset{x\to \infty }{\lim }{\displaystyle \frac{1}{x^2}}=0=\underset{x\to \infty }{\lim }-{\displaystyle \frac{1}{x^2}}\text{,}}$

and we conclude, using the Squeeze Theorem, that

${\displaystyle \underset{x\to \infty }{\lim }{\displaystyle \frac{\cos x}{x^2}}=0\text{.}}$

Example 3.76. By Theorem 3.29,

${\displaystyle \underset{x\to \infty }{\lim }{\displaystyle \frac{\pi x}{2x+1}}={\displaystyle \frac{\pi }{2}}\text{,}\text{and so}{\displaystyle \frac{\pi x}{2x+1}}\le {\displaystyle \frac{\pi }{2}}\text{for}x\to \infty }$

[To see this, take a large positive value of $x$ say 10,000.]

Hence, if

${\displaystyle t={\displaystyle \frac{\pi x}{2x+1}}\text{,}\text{then}t\to {\left({\displaystyle \frac{\pi }{2}}\right)}^{-}\text{as}x\to \infty \text{.}}$

By Theorem 3.20,

${\displaystyle \underset{x\to \infty }{\lim }tan\left({\displaystyle \frac{\pi x}{2x+1}}\right)=\underset{t\to {{\displaystyle \frac{\pi }{2}}}^{-}}{\lim }\tan t=\infty \text{.}}$

Remark 3.2

Polynomials have neither vertical nor horizontal asymptotes, because polynomials increase or decrease without bounds for either $x\to \infty$ or $x\to -\infty \text{.}$ Indeed, if

$P\left(x\right)=c_n{x}^n+c_{n-1}{x}^{n-1}+...+c_{1}x+c_0$

is a polynomial of degree $n$, then

${\displaystyle c_n{x}^n+c_{n-1}{x}^{n-1}+...+c_{1}x+c_0=x^n\left(c_n+{\displaystyle \frac{c_{n-1}}{x}}+...+{\displaystyle \frac{c_1}{x^{n-1}}}+{\displaystyle \frac{c_0}{x^n}}\right)}$

By Proposition 3.16, Example 3.65 and Theorem 3.18,

$\underset{x\to \pm \infty }{\lim }{c}_n{x}^n+c_{n-1}{x}^{n-1}+...+c_{1}x+c_0=\underset{x\to \pm \infty }{\lim }{c}_n{x}^n=\pm \infty \text{.}$

Example 3.77. Let us consider the cases below.

1. $\underset{x\to \infty }{\lim }\left(6-x^2\right)\left(3x^2+x-1\right)=\underset{x\to \infty }{\lim }-3x^4=-\infty$
2. $\underset{x\to -\infty }{\lim }3+4x^2+4x^3=\underset{x\to -\infty }{\lim }4x^3=-\infty$

Tutorial 4: The use of rationalization to evaluate a limit.

Tutorial 5: Application of several results to evaluate an infinite limit.