Chapter 2.2, Problem 33E

(a)

To determine

To find: The vertical asymptotes of the graph of $f\left(x\right)=\frac{\tan x}{\sin x}$ .

Answer to Problem 33E

The vertical asymptotes of the graph of $f\left(x\right)=\frac{\tan x}{\sin x}$ are $x=\frac{\pi }{2}+k\pi$ .

Explanation of Solution

Given information:The function is $f\left(x\right)=\frac{\tan x}{\sin x}$ .

Calculation:

Simplify the function.

  $\begin{array}{c}f\left(x\right)=\frac{\tan x}{\sin x}\\ =\frac{1}{\sin x}\times \frac{\sin x}{\cos x}\\ =\frac{1}{\cos x}\end{array}$

To graph a function $y=\frac{1}{\cos x}$ , follow the steps using graphing calculator.

First press “ON” button on graphical calculator, press $\boxed{\text{Y}=}$ key and enter right hand side of the equation $y=\frac{1}{\cos x}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{1}\boxed{÷}\boxed{(}\boxed{\text{COS}}\boxed{\text{X}}\boxed{)}\boxed{)}$ .

The display will show the equation,

  ${\text{Y}}_{\text{1}}=1/\left(\text{cosX}\right)$

Press the window key and adjust the window to $\left[-2\pi ,2\pi \right]$ by $\left[-3,3\right]$ .Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph of given function in standard window as shown below.

  

Figure (1)

As observed from graph of function $f\left(x\right)=\sec x$ , the vertical asymptotes to the graph are of type $x=\frac{\pi }{2}+k\pi$ for any integer $k$ .

Therefore, the vertical asymptotes of the graph of $f\left(x\right)=\frac{\tan x}{\sin x}$ are $x=\frac{\pi }{2}+k\pi$ .

(b)

To determine

To find:The behavior of the function $f\left(x\right)=\frac{\tan x}{\sin x}$ to the right and left of the vertical asymptotes.

Answer to Problem 33E

The value of $\underset{x\to {\left(\frac{\pi }{2}+k\pi \right)}^{-}}{\lim }f\left(x\right)$ , $\underset{x\to {\left(\frac{\pi }{2}+k\pi \right)}^{+}}{\lim }f\left(x\right)$ are $-\infty$ and $\infty$ respectively for $k$ being odd and $\infty$ and $-\infty$ respectively for $k$ being even.

Explanation of Solution

Given information:The function is $f\left(x\right)=\frac{\tan x}{\sin x}$ .

Calculation:

Firstly take the points of asymptote $x=\frac{\pi }{2}+k\pi$ for $k$ being odd integer.

As observed from the graph, the left hand limit for function at each point $x=\frac{\pi }{2}+k\pi$ is $-\infty$ .So, the value of $\underset{x\to {\left(\frac{\pi }{2}+k\pi \right)}^{-}}{\lim }f\left(x\right)$ is $-\infty$ .

As observed from the graph, the right hand limit for function at each point $x=\frac{\pi }{2}+k\pi$ is $\infty$ .So, the value of $\underset{x\to {\left(\frac{\pi }{2}+k\pi \right)}^{+}}{\lim }f\left(x\right)$ is $\infty$ .

Now take the points of asymptote $x=\frac{\pi }{2}+k\pi$ for $k$ being even integer.

As observed from the graph, the left hand limit for function at each point $x=\frac{\pi }{2}+k\pi$ is $\infty$ .So, the value of $\underset{x\to {\left(\frac{\pi }{2}+k\pi \right)}^{-}}{\lim }f\left(x\right)$ is $\infty$ .

As observed from the graph, the right hand limit for function at each point $x=\frac{\pi }{2}+k\pi$ is $-\infty$ .So, the value of $\underset{x\to {\left(\frac{\pi }{2}+k\pi \right)}^{+}}{\lim }f\left(x\right)$ is $-\infty$ .

Therefore, the value of $\underset{x\to {\left(\frac{\pi }{2}+k\pi \right)}^{-}}{\lim }f\left(x\right)$ , $\underset{x\to {\left(\frac{\pi }{2}+k\pi \right)}^{+}}{\lim }f\left(x\right)$ are $-\infty$ and $\infty$ respectively for $k$ being odd and $\infty$ and $-\infty$ respectively for $k$ being even.