# The limit of each of the given functions and explain if the limit does not exist. ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

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.

Expert Solution

(i) The value of limx2+f(x)=3.

(ii) The value of limx3+f(x)=0.

(iii) The value of limx3f(x) does not exist.

(iv) The value of limx4f(x)=2

(v) The value of limx0f(x) is infinity.

(vi) The value of limx2f(x) is

(vii) The value of limxf(x)=4.

(viii) The value of limxf(x)=1

### Explanation of Solution

Calculation:

Section (i)

Obtain the value of limx2+f(x).

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, limx2+f(x)=3.

Section (ii)

Obtain the value of limx3+f(x).

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, limx3+f(x)=0.

Section (iii)

Obtain the value of limx3f(x).

From section (ii),

limx3+f(x)=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, limx3+f(x)=2

Recall the definition that limxaf(x) exists only when limxaf(x)=limxa+f(x).

Since limx3+f(x)limx3f(x)

Therefore, limx3f(x) does not exists.

Section (iv)

Obtain the value of limx4f(x)

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, limx4f(x)=2.

Section (v)

Obtain the value of limx0f(x)

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, limx0f(x)=.

Section (vi)

Obtain the value of limx2f(x)

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

Therefore,limx2f(x)=

Section (vii)

Obtain the value of limxf(x)

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

Therefore,limxf(x)=4

Section (viii)

Obtain the value of limxf(x)

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

Therefore,limxf(x)=1

(b)

To determine

### To state: The equation of horizontal asymptotes.

Expert Solution

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 limxf(x)=c or limxf(x)=c

From the previous parts,

limxf(x)=4 and limxf(x)=1

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

(c)

To determine

### To state: The equation of vertical asymptotes.

Expert Solution

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 limxaf(x)=± or limxa+f(x)=±

From the previous parts,

limx0f(x)= and limx2f(x)=

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

(d)

To determine

### To find: The numbers at which f is discontinuous.

Expert Solution

f is discontinuous at x=3,0,2 and 4.

### Explanation of Solution

From the given graph, limx3f(x)limx3+f(x)

Thus, f is discontinuous at x=3.

From the given graph, limx0f(x)=

Thus, f is discontinuous at x=0

From the given graph, limx2f(x)=

Thus, f is discontinuous at x=2.

From the given graph, limx4f(x)=2 and f(4)=1

Recall from the definition that f is continuous at x=a if limxaf(x)=f(a).

Thus, f is discontinuous at x=4.

Therefore, f is discontinuous at x=3,0,2 and 4.

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