Chapter 12.2, Problem 6E

a.

To determine

To find: To find each limit visually from the given graph.

Answer to Problem 6E

The limit for $\underset{x\to 1}{\lim }f\left(x\right)\text{is}0$ .

Explanation of Solution

Given:

  

From the given graph we can observe that $f\left(x\right)$ approaches towards $0$ as $x\to 1$ from both negative and positive directions. It has been defined exactly at $0\text{at}x=1$ . Direct substitution technique can be used to verify it because $f\left(x\right)$ is a rational function with a non-zero denominator for $x=1$ .

Therefore we can conclude that $\underset{x\to 1}{\lim }f\left(x\right)\text{is}0$ .

b.

To determine

To find: To find each limit visually from the given graph.

Answer to Problem 6E

The limit for $\underset{x\to 2}{\lim }f\left(x\right)\text{is}1$ .

Explanation of Solution

Given:

  

From the given graph we can observe that $f\left(x\right)$ approaches towards $1$ as $x\to 2$ from both negative and positive directions. It has been defined exactly at $1\text{at}x=2$ . Direct substitution technique can be used to verify it because $f\left(x\right)$ is a rational function with a non-zero denominator for $x=2$ .

Therefore we can conclude that $\underset{x\to 2}{\lim }f\left(x\right)\text{is}1$ .

c.

To determine

To find: To find each limit visually from the given graph.

Answer to Problem 6E

The limit for $\underset{x\to -1}{\lim }g\left(x\right)\text{is}-2$ .

Explanation of Solution

Given:

  

From the given graph we can observe that $f\left(x\right)$ approaches towards $-2$ as $x\to -1$ from both negative and positive directions.

Therefore we can conclude that $\underset{x\to 2}{\lim }f\left(x\right)\text{is}-2$ .

d.

To determine

To find: To identify another function that agrees with the given function at all but one point.

Answer to Problem 6E

A function that agrees with the given function at all but one point is $\left(x-1\right)$ .

Explanation of Solution

Given:

  

To find another function that agrees with the given function at all but one point we have to use the dividing out technique. Factor the numerator and divide out any common factors between the numerator and denominator.

  $\begin{array}{l}\frac{x^2-1}{x+1}=\frac{\left(x-1\right)\left(x+1\right)}{x+1}\\ \frac{x^2-1}{x+1}=x-1\end{array}$

Thus we can find another function that agrees with the given function at all but one point.