Chapter 11.2, Problem 6E

(a)

To determine

To find: the $\underset{x\to 0}{\lim }g\left(x\right)$ by using the graph and identify another function at all but one point.

Answer to Problem 6E

  $\underset{x\to 0}{\lim }g\left(x\right)=1$ .

Explanation of Solution

Given:

  

Concept used:

Describe the behaviour of $g\left(x\right)$ to the immediate left of $x=a$ .

The left-hand limit at $x=a$ is:

  $\underset{x\to a^{-}}{\lim }g\left(x\right)$ .

Describe the behaviour of $g\left(x\right)$ to the immediate right of $x=a$ .

The right-hand limit at $x=a$ is:

  $\underset{x\to a^{+}}{\lim }g\left(x\right)$ .

Give the precise value of $g\left(x\right)$ take at $x=a$ .

Calculation:

As it can see from the graph, $g\left(x\right)\to 1\text{ as }x\to 0$ from both the positive and negative directions, therefore it can say that $\underset{x\to 0}{\lim }g\left(x\right)=1$ .

Hence, $\underset{x\to 0}{\lim }g\left(x\right)=1$ .

(b)

To determine

To find: the $\underset{x\to -1}{\lim }g\left(x\right)$ by using the graph and identify another function at all but one point.

Answer to Problem 6E

  $\underset{x\to -1}{\lim }g\left(x\right)=3$ .

Explanation of Solution

Given:

  

Concept used:

Describe the behaviour of $g\left(x\right)$ to the immediate left of $x=a$ .

The left-hand limit at $x=a$ is:

  $\underset{x\to a^{-}}{\lim }g\left(x\right)$ .

Describe the behaviour of $g\left(x\right)$ to the immediate right of $x=a$ .

The right-hand limit at $x=a$ is:

  $\underset{x\to a^{+}}{\lim }g\left(x\right)$ .

Give the precise value of $g\left(x\right)$ take at $x=a$ .

Calculation:

As it can see from the graph, $g\left(x\right)\to 3\text{ as }x\to -1$ from both the positive and negative direction also, it can find this direct substitution, since, $g\left(x\right)$ is a rational function with a non- zero denominator for $x=-1$ .

Hence, $\underset{x\to -1}{\lim }g\left(x\right)=3$ .

(c)

To determine

To find: the $\underset{x\to -2}{\lim }g\left(x\right)$ by using the graph and identify another function at all but one point.

Answer to Problem 6E

  $\underset{x\to -2}{\lim }g\left(x\right)=5$ .

Explanation of Solution

Given:

  

Concept used:

Describe the behaviour of $g\left(x\right)$ to the immediate left of $x=a$ .

The left-hand limit at $x=a$ is:

  $\underset{x\to a^{-}}{\lim }g\left(x\right)$ .

Describe the behaviour of $g\left(x\right)$ to the immediate right of $x=a$ .

The right-hand limit at $x=a$ is:

  $\underset{x\to a^{+}}{\lim }g\left(x\right)$ .

Give the precise value of $g\left(x\right)$ take at $x=a$ .

Calculation:

It can also see that $g\left(x\right)\to 5\text{ as }x\to -2$ from both directions, and it appears as if it is also defined to be exactly $5$ at $x=-2$ . It can verify this by direct substitution, since $g\left(x\right)$ is a rational function with a non-zero denominator for $x=-2$ .

It can conclude that $\underset{x\to -2}{\lim }g\left(x\right)=5$ .

(d)

To determine

To find: the $\underset{x\to 1}{\lim }g\left(x\right)$ by using the graph and identify another function at all but one point.

Answer to Problem 6E

  $\underset{x\to 1}{\lim }g\left(x\right)$ does not exist.

Explanation of Solution

Given:

  

Concept used:

In order to find another function that agrees with $g\left(x\right)$ at all but one point, it use the dividing out technique, described at the beginning of the chapter. factor the numerator then divide out any common factors between the numerator and denominator.

Indeterminate form:

  $g\left(0\right)=\frac{0}{0}$ .

Calculation:

In order to find another function that agrees with $g\left(x\right)$ at all but one point, it use the dividing out technique, described at the beginning of the chapter. factor the numerator then divide out any common factors between the numerator and denominator.

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

Also, it is noticed that technique can use to find the limit from part $\left(a\right)$ .

Since, $g\left(0\right)=\frac{0}{0}$ which is indeterminate, but by plugging $x=0$ into this function that agrees with $g\left(x\right)$ at all but one point, it can find the limits.

Hence, $\underset{x\to 1}{\lim }g\left(x\right)$ does not exist.