Chapter 13.4, Problem 3E

a.

To determine

To find : The following limit i) $\underset{x\to \infty }{\lim }f\left(x\right)$ and ii) $\underset{x\to -\infty }{\lim }f\left(x\right)$

Answer to Problem 3E

  $\underset{x\to \infty }{\lim }f\left(x\right)=-1$ and $\underset{x\to -\infty }{\lim }f\left(x\right)=2$

Explanation of Solution

Given information:

The graph of the function f is given.

Calculation :

i)

Let $f$ be a function defined on some interval $\left(a,\infty \right)$ . Then

  $\underset{x\to \infty }{\lim }f\left(x\right)=L$

Means that the value of $f\left(x\right)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently large.

Therefore, from graph it can be observed that:

  $\underset{x\to \infty }{\lim }f\left(x\right)=-1$

ii)

If $f$ be a function defined on some interval $\left(-\infty ,a\right)$ . Then

  $\underset{x\to -\infty }{\lim }f\left(x\right)=L$

Means that the value of $f\left(x\right)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently large negative.

Therefore, from graph it can be observed that:

  $\underset{x\to -\infty }{\lim }f\left(x\right)=2$

Hence,

  $\underset{x\to \infty }{\lim }f\left(x\right)=-1$ and $\underset{x\to -\infty }{\lim }f\left(x\right)=2$

b.

To determine

To state : The equation of the horizontal asymptotes.

Answer to Problem 3E

The horizontal asymptotes are $y=-1$ and $y=2$.

Explanation of Solution

Given information:

The graph of the function f is given.

Calculation :

The line $y=L$ is called the horizontal asymptotes of the curve $y=f\left(x\right)$ if either

  $\underset{x\to \infty }{\lim }f\left(x\right)=L$ or $\underset{x\to -\infty }{\lim }f\left(x\right)=L$.

From part (a) $\underset{x\to \infty }{\lim }f\left(x\right)=-1$ and $\underset{x\to -\infty }{\lim }f\left(x\right)=2$.

Therefore, horizontal asymptotes are

  $y=-1$ and $y=2$.

Hence,

The horizontal asymptotes are $y=-1$ and $y=2$.