Chapter 2, Problem 1RE

(a)

To determine

To find: The limit of each of the given functions and explain if the limit does not exist.

Answer to Problem 1RE

(i) The value of $\underset{x\to 2^{+}}{\lim }f\left(x\right)=3$.

(ii) The value of $\underset{x\to -3^{+}}{\lim }f\left(x\right)=0$.

(iii) The value of $\underset{x\to -3}{\lim }f\left(x\right)$ does not exist.

(iv) The value of $\underset{x\to 4}{\lim }f\left(x\right)=2$

(v) The value of $\underset{x\to 0}{\lim }f\left(x\right)$ is infinity.

(vi) The value of $\underset{x\to 2^{-}}{\lim }f\left(x\right)$ is $-\infty$

(vii) The value of $\underset{x\to \infty }{\lim }f\left(x\right)=4$.

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

Explanation of Solution

Calculation:

Section (i)

Obtain the value of $\underset{x\to 2^{+}}{\lim }f\left(x\right)$.

From the given graph, it is found that the curve move towards $y=3$ as x approaches 2 from the right, that is $x>2$.

Therefore, $\underset{x\to 2^{+}}{\lim }f\left(x\right)=3$.

Section (ii)

Obtain the value of $\underset{x\to -3^{+}}{\lim }f\left(x\right)$.

From the given graph, it is found that the curve move towards $y=0$ as x approaches $-3$ from the right, that is $x>-3$.

Therefore, $\underset{x\to -3^{+}}{\lim }f\left(x\right)=0$.

Section (iii)

Obtain the value of $\underset{x\to -3}{\lim }f\left(x\right)$.

From section (ii),

$\underset{x\to -3^{+}}{\lim }f\left(x\right)=0$

From the given graph, it is found that the curve move towards $y=-2$ as x approaches $-3$ from the left, that is $x<-3$.

Thus, $\underset{x\to -3^{+}}{\lim }f\left(x\right)=-2$

Recall the definition that $\underset{x\to a}{\lim }f\left(x\right)$ exists only when $\underset{x\to a^{-}}{\lim }f\left(x\right)=\underset{x\to a^{+}}{\lim }f\left(x\right)$.

Since $\underset{x\to -3^{+}}{\lim }f\left(x\right)\ne \underset{x\to -3^{-}}{\lim }f\left(x\right)$

Therefore, $\underset{x\to -3}{\lim }f\left(x\right)$ does not exists.

Section (iv)

Obtain the value of $\underset{x\to 4}{\lim }f\left(x\right)$

From the given graph, it is found that the curve move towards $y=2$ as x approaches $4$ from both the left and right side.

Therefore, $\underset{x\to 4}{\lim }f\left(x\right)=2$.

Section (v)

Obtain the value of $\underset{x\to 0}{\lim }f\left(x\right)$

From the given graph, it is found that the curve move towards infinity as x approaches $0$ from both the left and right side.

Therefore, $\underset{x\to 0}{\lim }f\left(x\right)=\infty$.

Section (vi)

Obtain the value of $\underset{x\to 2^{-}}{\lim }f\left(x\right)$

From the given graph, it is found that the curve move towards negative infinity as x approaches 2 from the left side.

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

Section (vii)

Obtain the value of $\underset{x\to \infty }{\lim }f\left(x\right)$

From the given graph, it is found that the curve move towards $y=4$ as x approaches infinity.

Therefore,$\underset{x\to \infty }{\lim }f\left(x\right)=4$

Section (viii)

Obtain the value of $\underset{x\to -\infty }{\lim }f\left(x\right)$

From the given graph, it is found that the curve move towards $y=-1$ as x approaches negative infinity.

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

(b)

To determine

To state: The equation of horizontal asymptotes.

Answer to Problem 1RE

The equation of horizontal asymptotes are $y=4$ and $y=-1$.

Explanation of Solution

Recall from the definition that $y=c$ is horizontal asymptote if $\underset{x\to -\infty }{\lim }f\left(x\right)=c$ or $\underset{x\to \infty }{\lim }f\left(x\right)=c$

From the previous parts,

$\underset{x\to \infty }{\lim }f\left(x\right)=4$ and $\underset{x\to -\infty }{\lim }f\left(x\right)=-1$

Thus, equation of horizontal asymptotes are $y=4$ and $y=-1$.

(c)

To determine

To state: The equation of vertical asymptotes.

Answer to Problem 1RE

The equation of vertical asymptotes are $x=0$ and $x=2$.

Explanation of Solution

Recall from the definition that $x=a$ is vertical asymptote if $\underset{x\to a^{-}}{\lim }f\left(x\right)=\pm \infty$ or $\underset{x\to a^{+}}{\lim }f\left(x\right)=\pm \infty$

From the previous parts,

$\underset{x\to 0}{\lim }f\left(x\right)=\infty$ and $\underset{x\to 2^{-}}{\lim }f\left(x\right)=-\infty$

Thus, equation of vertical asymptotes are $x=0$ and $x=2$.

(d)

To determine

To find: The numbers at which f is discontinuous.

Answer to Problem 1RE

f is discontinuous at $x=-3,0,2\text{ and 4}$.

Explanation of Solution

From the given graph, $\underset{x\to 3^{-}}{\lim }f\left(x\right)\ne \underset{x\to 3^{+}}{\lim }f\left(x\right)$

Thus, f is discontinuous at $x=-3$.

From the given graph, $\underset{x\to 0}{\lim }f\left(x\right)=\infty$

Thus, f is discontinuous at $x=0$

From the given graph, $\underset{x\to 2^{-}}{\lim }f\left(x\right)=-\infty$

Thus, f is discontinuous at $x=2$.

From the given graph, $\underset{x\to 4}{\lim }f\left(x\right)=2$ and $f\left(4\right)=1$

Recall from the definition that f is continuous at $x=a$ if $\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)$.

Thus, f is discontinuous at $x=4$.

Therefore, f is discontinuous at $x=-3,0,2\text{ and 4}$.