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Activity2.8Vertical Asymptotes

Whenever \(\lim\limits_{x\to a}\fe{f}{x}\neq0\) but \(\lim\limits_{x\to a}\fe{g}{x}=0\), then \(\lim\limits_{x\to a}\frac{\fe{f}{x}}{\fe{g}{x}}\) does not exist because from either side of \(a\) the value of \(\frac{\fe{f}{x}}{\fe{g}{x}}\) has an absolute value that will become arbitrarily large. In these situations the line \(x=a\) is a vertical asymptote for the graph of \(y=\frac{\fe{f}{x}}{\fe{g}{x}}\). For example, the line \(x=2\) is a vertical asymptote for the function \(h\) defined by \(\fe{h}{x}=\frac{x+5}{2-x}\). We say that \(\lim\limits_{x\to 2}\frac{x+5}{2-x}\) has the form “not-zero over zero.” (Specifically, the form of \(\lim\limits_{x\to 2}\frac{x+5}{2-x}\) is \(\frac{7}{0}\).) Every limit with form “not-zero over zero” does not exist. However, we frequently can communicate the non-existence of the limit using an infinity symbol. In the case of \(\fe{h}{x}=\frac{x+5}{2-x}\) it's pretty easy to see that \(\fe{h}{1.99}\) is a positive number whereas \(\fe{h}{2.01}\) is a negative number. Consequently, we can infer that \(\lim\limits_{x\to 2^{-}}\fe{h}{x}=\infty\) and \(\lim\limits_{x\to 2^{+}}\fe{h}{x}=-\infty\). Remember, these equations are communicating that the limits do not exist as well as the reason for their non-existence. There is no short-hand way to communicate the non-existence of the two-sided limit \(\lim\limits_{x\to 2}\fe{h}{x}\).

Subsection2.8.1Exercises

Suppose that \(\fe{g}{t}=\frac{t+4}{t+3}\).

1

What is the vertical asymptote on the graph of \(y=\fe{g}{t}\)?

2

Write an equality about \(\lim\limits_{t\to-3^{-}}\fe{g}{t}\).

3

Write an equality about \(\lim\limits_{t\to-3^{+}}\fe{g}{t}\).

4

Is it possible to write an equality about \(\lim\limits_{t\to-3}\fe{g}{t}\)? If so, do it.

5

Which of the following limits exist? \begin{equation*}\lim\limits_{t\to-3^{-}}\fe{g}{t}\text{?}\quad\lim\limits_{t\to-3^{+}}\fe{g}{t}\text{?}\quad\lim\limits_{t\to-3}\fe{g}{t}\text{?}\end{equation*}

Suppose that \(\fe{z}{x}=\frac{7-3x^2}{\left(x-2\right)^2}\).

6

What is the vertical asymptote on the graph of \(y=\fe{z}{x}\)?

7

Is it possible to write an equality about \(\lim\limits_{x\to2}\fe{z}{x}\)? If so, do it.

8

What is the horizontal asymptote on the graph of \(y=\fe{z}{x}\)?

9

Which of the following limits exist? \begin{equation*}\lim\limits_{x\to2^{-}}\fe{z}{x}\text{?}\quad\lim\limits_{x\to2^{+}}\fe{z}{x}\text{?}\quad\lim\limits_{x\to2}\fe{z}{x}\text{?}\end{equation*}

10

Consider the function \(f\) defined by \(\fe{f}{x}=\frac{x+7}{x-8}\). Complete Table 2.8.1 without the use of your calculator.

Use this as an opportunity to discuss why limits of form “not-zero over zero” are “infinite limits.” What limit equation is being illustrated in the table?

\(x\) \(x+7\) \(x-8\) \(\fe{f}{x}\)
\(8.1\) \(15.1\) \(0.1\)
\(8.01\) \(15.01\) \(0.01\)
\(8.001\) \(15.001\) \(0.001\)
\(8.0001\) \(15.0001\) \(0.0001\)
Table2.8.1\(\fe{f}{x}=\frac{x+7}{x-8}\)

Hear me, and hear me loud…\(\infty\) does not exist. This, in part, is why we cannot apply Limit Law R1 to an expression like \(\lim\limits_{x\to\infty}x\) to see that \(\lim\limits_{x\to\infty}x=\infty\). When we write, say, \(\lim\limits_{x\to7}x=7\), we are replacing the limit, we are replacing the limit expression with its value—that's what the replacement laws are all about! When we write \(\lim\limits_{x\to\infty}x=\infty\), we are not replacing the limit expression with a value! We are explicitly saying that the limit has no value (i.e. does not exist) as well as saying the reason the limit does not exist. The limit laws can only be applied when all of the limits in the equation exist. With this in mind, discuss and decide whether each of the following equations are true or false.

11

True or False? \(\lim\limits_{x\to0}\left(\dfrac{e^x}{e^x}\right)=\dfrac{\lim\limits_{x\to0}e^x}{\lim\limits_{x\to0}e^x}\)

12

True or False? \(\lim\limits_{x\to1}\dfrac{e^x}{\fe{\ln}{x}}=\dfrac{\lim\limits_{x\to1}e^x}{\lim\limits_{x\to1}\fe{\ln}{x}}\)

13

True or False? \(\lim\limits_{x\to0^{+}}\left(2\fe{\ln}{x}\right)=2\lim\limits_{x\to0^{+}}\fe{\ln}{x}\)

14

True or False? \(\lim\limits_{x\to\infty}\left(e^x-\fe{\ln}{x}\right)=\lim\limits_{x\to\infty}e^x-\lim\limits_{x\to\infty}\fe{\ln}{x}\)

15

True or False? \(\lim\limits_{x\to-\infty}\left(\dfrac{e^{-x}}{e^{-x}}\right)=\dfrac{\lim\limits_{x\to-\infty}e^{-x}}{\lim\limits_{x\to-\infty}e^{-x}}\)

16

True or False? \(\lim\limits_{x\to1}\left(\dfrac{\fe{\ln}{x}}{e^x}\right)=\dfrac{\lim\limits_{x\to1}\fe{\ln}{x}}{\lim\limits_{x\to1}e^x}\)

17

True or False? \(\lim\limits_{\theta\to\infty}\dfrac{\fe{\sin}{\theta}}{\fe{\sin}{\theta}}=\dfrac{\lim\limits_{\theta\to\infty}\fe{\sin}{\theta}}{\lim\limits_{\theta\to\infty}\fe{\sin}{\theta}}\)

18

True or False? \(\lim\limits_{x\to-\infty}e^{\sfrac{1}{x}}=e^{\lim_{x\to-\infty}\sfrac{1}{x}}\)

19

Mindy tried to evaluate \(\lim\limits_{x\to6^{+}}\frac{4x-24}{x^2-12x+36}\) using the limit laws. Things went horribly wrong for Mindy (her work is shown below). Identify what is wrong in Mindy's work and discuss what a more reasonable approach might have been. This “solution” is not correct! Do not emulate Mindy's work! \begin{align*} \lim_{x\to6^{+}}\frac{4x-24}{x^2-12x+36}&=\lim_{x\to6^{+}}\frac{4(x-6)}{(x-6)^2}\\ &=\lim_{x\to6^{+}}\frac{4}{x-6}\\ &=\frac{\lim\limits_{x\to6^{+}}4}{\lim\limits_{x\to6^{+}}(x-6)}&&\knowl{./knowl/lla5.html}{\text{Limit Law A5}}\\ &=\frac{\lim\limits_{x\to6^{+}}4}{\lim\limits_{x\to6^{+}}x-\lim\limits_{x\to6^{+}}6}&&\knowl{./knowl/lla2.html}{\text{Limit Law A2}}\\ &=\frac{4}{6-6}&&\knowl{./knowl/llr1.html}{\text{Limit Law R1}}, \knowl{./knowl/llr2.html}{\text{Limit Law R2}}\\ &=\frac{4}{0} \end{align*} This “solution” is not correct! Do not emulate Mindy's work!