Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 5

Applications of the Derivative

Asymptotes

In Unit 3, we defined and learned how to find the vertical and horizontal asymptotes of a function (see Definitions 3.19 and 3.28). As we mentioned in the previous section, the domain of a function indicates its possible vertical asymptotes. Let us review the procedure for vertical asymptotes once more.

Example 5.20. In Example 5.2, we decided that the function

${\displaystyle h\left(x\right)={\displaystyle \frac{4x-5}{4x^2-6x}}}$

may have a vertical asymptotes at ${\displaystyle x=0}$ or ${\displaystyle x=3∕2}$.

To see if it does so, we take any one of the limits listed in Definition 3.19. We find that

${\displaystyle \underset{x\to 0^{+}}{lim}{\displaystyle \frac{4x-5}{4x^2-6x}}=\underset{x\to 0^{+ }}{lim}\left({\displaystyle \frac{4x-5}{4x-6}}\right)\left({\displaystyle \frac{1}{x}}\right)=\infty \text{,}}$

since

${\displaystyle \underset{x\to 0^{+}}{lim}{\displaystyle \frac{4x-5}{4x-6}}={\displaystyle \frac{5}{6}}>0\text{and}\underset{x\to 0^{+}}{lim}{\displaystyle \frac{1}{x}}=\infty \text{.}}$

Furthermore,

${\displaystyle \underset{x\to 3∕2^{+}}{lim}{\displaystyle \frac{4x-5}{4x^2-6x}}=\underset{x\to 3∕2^{+ }}{lim}\left({\displaystyle \frac{1}{2x-3}}\right)\left({\displaystyle \frac{4x-5}{2x}}\right)=\infty \text{,}}$

since

${\displaystyle \underset{x\to 3∕2^{+}}{lim}{\displaystyle \frac{4x-5}{2x}}={\displaystyle \frac{1}{3}}>0\text{and}\underset{x\to 3∕2^{+}}{lim}{\displaystyle \frac{1}{2x-3}}=\infty \text{.}}$

Hence, the function has vertical asymptotes at ${\displaystyle x=0}$ and ${\displaystyle x=3∕2}$.

Example 5.21. In Example 5.3, we saw that ${\displaystyle x=0}$ and ${\displaystyle x=7∕2}$ may be vertical asymptotes of the function

${\displaystyle g\left(x\right)={\displaystyle \frac{x^2+1}{x{\left(2x-7\right)}^2}}\text{.}}$

Taking limits, we find that

${\displaystyle \underset{x\to 0^{+}}{lim}{\displaystyle \frac{x^2+1}{x{\left(2x-7\right)}^2}}=\underset{x\to 0^{+ }}{lim}\left({\displaystyle \frac{x^2+1}{{\left(2x-7\right)}^2}}\right)\left({\displaystyle \frac{1}{x}}\right)=\infty \text{,}}$

since

${\displaystyle \underset{x\to 0^{+}}{lim}{\displaystyle \frac{x^2+1}{{\left(2x-7\right)}^2}}={\displaystyle \frac{1}{49}}>0\text{and}\underset{x\to 0^{+}}{lim}{\displaystyle \frac{1}{x}}=\infty ;}$

and

${\displaystyle \underset{x\to 7∕2^{+}}{lim}{\displaystyle \frac{x^2+1}{x{\left(2x-7\right)}^2}}=\underset{x\to 7∕2^{+ }}{lim}\left({\displaystyle \frac{x^2+1}{x}}\right)\left({\displaystyle \frac{1}{{\left(2x-7\right)}^2}}\right)=\infty \text{,}}$

since

${\displaystyle \underset{x\to 7∕2^{+}}{lim}{\displaystyle \frac{x^2+1}{x}}={\displaystyle \frac{53}{14}}>0\text{and}\underset{x\to 7∕2^{+}}{lim}{\displaystyle \frac{1}{{\left(2x-7\right)}^2}}=\infty \text{.}}$

Hence, the function has vertical asymptotes at ${\displaystyle x=0}$ and ${\displaystyle x=7∕2}$.

Example 5.22. Consider the function

${\displaystyle f\left(x\right)=\text{tan}\left(2x+1\right)}$

(see Example 5.6.

This function is periodic: its graph repeats itself every interval

${\displaystyle \left({\displaystyle \frac{\left(2k+1\right)\pi -2}{4}},{\displaystyle \frac{\left(2k+3\right)\pi -2}{4}}\right)\text{.}}$

Hence, it is enough to check if the function has a vertical asymptote at the endpoints of the open interval for ${\displaystyle k=0\text{.}}$

Taking limits, we see that

${\displaystyle \text{if }t=2x+1\text{then}t\to {{\displaystyle \frac{\pi }{2}}}^{+}\text{as}x\to {\left({\displaystyle \frac{\pi -2}{4}}\right)}^{+}\text{.}}$

Hence,

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

We can check in a similar way that

${\displaystyle \underset{x\to {\left({\displaystyle \frac{3\pi -2}{4}}\right)}^{+}}{lim}\text{tan}\left(2x+1\right)=\underset{t\to {\left({\displaystyle \frac{3\pi }{2}}\right)}^{+}}{lim}\text{tan }t=-\infty \text{.}\text{ [Try it.]}}$

Therefore, the function has vertical asymptotes at each endpoint of the open intervals of its domain.

Example 5.23. The function

${\displaystyle {\displaystyle \frac{x^2+8x+15}{x^2+5x+6}}}$

may have vertical asymptotes at ${\displaystyle x=-3}$ and ${\displaystyle x=-2}$. [Why?]

Taking limits, we find

${\displaystyle \underset{x\to -3^{-}}{lim}{\displaystyle \frac{x^2+8x+15}{x^2+5x+6}}=\underset{x\to -3^{-}}{lim}{\displaystyle \frac{x+5}{x+2}}=-2}$

and

${\displaystyle \underset{x\to -2^{-}}{lim}{\displaystyle \frac{x^2+8x+15}{x^2+5x+6}}=\underset{x\to -2^{-}}{lim}{\displaystyle \frac{x+5}{x+2}}=-\infty \text{.}}$

Hence, ${\displaystyle x=-2}$ is the only vertical asymptote for this function.

A line ${\displaystyle y=mx+b}$ with ${\displaystyle m\ne 0}$ is a slant asymptote of a function $f(x)$ if

${\displaystyle \underset{x\to \pm \infty }{lim}f\left(x\right)-\left(mx+b\right)=0\text{.}}$

In this course, we only consider the case of slant asymptotes of rational functions in which the degree of the numerator is one more than the degree of the denominator. See Example 4 on page 237 of the textbook.