Chapter 2, Problem 6RCC

(a)

To determine

To find: Whether the given curve has vertical asymptotes or horizontal asymptotes.

Answer to Problem 6RCC

The given curve has no asymptotes.

Explanation of Solution

Result used:

Definition of vertical asymptote:

The vertical asymptote of the function $y=f\left(x\right)$ is $x=a$, if y approaches $\pm \infty$ as x approaches a from the right or left.

Limit is defined as $\underset{x\to a^{-}}{\lim }f\left(x\right)=\pm \infty$(or)$\underset{x\to a^{+}}{\lim }f\left(x\right)=\pm \infty$ (or) both.

Definition of horizontal asymptote:

The horizontal asymptote of the function $y=f\left(x\right)$ is $y=b$, if y approaches b as x approaches $\pm \infty$.

Limit is defined as $\underset{x\to -\infty }{\lim }f\left(x\right)=b$(or)$\underset{x\to +\infty }{\lim }f\left(x\right)=b$(or) both.

Graph:

The graph of a function $y=x^4$

Calculation:

For vertical asymptotes:

$\begin{array}{c}\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }{x}^4\\ ={\left(0\right)}^4\\ =0\end{array}$

There are no vertical asymptotes.

For horizontal asymptotes:

$\begin{array}{c}\underset{x\to \pm \infty }{\lim }f\left(x\right)=\underset{x\to \pm \infty }{\lim }{x}^4\\ ={\left(\pm \infty \right)}^4\\ =\infty \end{array}$

There are no horizontal asymptotes.

(b)

To determine

To find: Whether the curve y = sin x have vertical asymptotes or horizontal asymptotes.

Answer to Problem 6RCC

The graph of y = sin x has no asymptotes.

Explanation of Solution

The graph of a function $y=\sin \left(x\right)$

There are no vertical asymptotes, because the function $y=\sin \left(x\right)$ is defined for all real numbers so there are no vertical lines.

For vertical asymptotes:

$\begin{array}{c}\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }\sin \left(x\right)\\ =\sin \left(0\right)\\ =0\end{array}$

There are no vertical asymptotes.

For horizontal asymptotes:

There are no horizontal asymptotes, because from the graph the function $y=\sin \left(x\right)$ oscillates between –1 and 1 so there are no horizontal lines. The limit of the function is

$\begin{array}{c}\underset{x\to \pm \infty }{\lim }f\left(x\right)=\underset{x\to \pm \infty }{\lim }\sin \left(x\right)\\ \underset{x\to \pm \infty }{\lim }f\left(x\right)=\sin \left(\infty \right)\in \left[-1,1\right]\end{array}$

(c)

To determine

To find: Whether the curve y = tan x have vertical asymptotes or horizontal asymptotes.

Answer to Problem 6RCC

There are only vertical asymptotes at $x=\left(2n+1\right)\frac{\pi }{2}$.

Explanation of Solution

Graph:

The graph of a function $y=\tan \left(x\right)$.

Calculation:

For horizontal asymptotes,

$\begin{array}{c}\underset{x\to \pm \infty }{\lim }f\left(x\right)=\underset{x\to \pm \infty }{\lim }\tan \left(x\right)\\ \underset{x\to \pm \infty }{\lim }f\left(x\right)=\tan \left(\infty \right)\end{array}$

There are no horizontal asymptotes.

From the graph there are vertical asymptotes at $x=\left(2n+1\right)\frac{\pi }{2}$, n = 0,1,2,3,…

(d)

To determine

To find: Whether the given curve has vertical asymptotes or horizontal asymptotes

Answer to Problem 6RCC

There are only horizontal asymptotes at $y=\frac{\pi }{2}$ and $y=-\frac{\pi }{2}$.

Explanation of Solution

Graph:

The graph of a function $y={\tan }^{-1}\left(x\right)$.

Calculation:

For vertical asymptotes,

$\begin{array}{c}\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }{\tan }^{-1}\left(x\right)\\ ={\tan }^{-1}\left(0\right)\\ =0\end{array}$

There are no vertical asymptotes.

From the graph there are horizontal asymptotes at $y=\frac{\pi }{2}$ and $y=-\frac{\pi }{2}$.

(e)

To determine

To find: Whether the given curve has vertical asymptotes or horizontal asymptotes

Answer to Problem 6RCC

There are only horizontal asymptote at $y=0$.

Explanation of Solution

Graph:

The graph of a function $y=e^x$

Calculation:

For vertical asymptotes,

$\begin{array}{c}\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }{e}^x\\ =e^0\\ =1\end{array}$

There are no vertical asymptotes.

From the graph there are horizontal asymptote at $y=0$.

(f)

To determine

To find: Whether the given curve has vertical asymptotes or horizontal asymptotes.

Answer to Problem 6RCC

There are only vertical asymptote at $x=0$.

Explanation of Solution

Graph:

The graph of a function $y=\ln \left(x\right)$

Calculation:

For horizontal asymptotes,

$\begin{array}{c}\underset{x\to \pm \infty }{\lim }f\left(x\right)=\underset{x\to \pm \infty }{\lim }\ln \left(x\right)\\ =\ln \left(\infty \right)\\ =\infty \end{array}$

There are no horizontal asymptotes

From the graph there is vertical asymptote at $x=0$.

(g)

To determine

To find: Whether the given curve has vertical asymptotes or horizontal asymptotes

Answer to Problem 6RCC

There is vertical asymptote at $x=0$.

There is horizontal asymptote at $y=0$.

Explanation of Solution

Graph:

The graph of a function $y=\frac{1}{x}$

Calculation:

There is horizontal asymptote at $y=0$.

That is $\underset{x\to \infty }{\lim }\left(\frac{1}{x}\right)=0$.

There is vertical asymptote at $x=0$.

That is $\underset{x\to 0^{+}}{\lim }\left(\frac{1}{x}\right)=\infty$ and $\underset{x\to 0^{-}}{\lim }\left(\frac{1}{x}\right)=-\infty$.

(h)

To determine

To find: Whether the given curve has vertical asymptotes or horizontal asymptotes.

Answer to Problem 6RCC

The given curve has no asymptotes.

Explanation of Solution

Graph:

The graph of a function $y=\sqrt{x}$

Calculation:

For vertical asymptotes,

$\begin{array}{c}\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }\sqrt{x}\\ =\sqrt{0}\\ =0\end{array}$

There are no vertical asymptotes.

For horizontal asymptotes:

$\begin{array}{c}\underset{x\to \pm \infty }{\lim }f\left(x\right)=\underset{x\to \pm \infty }{\lim }\sqrt{x}\\ =\sqrt{\infty }\\ =\infty \end{array}$

There are no horizontal asymptotes.