One-Sided Limits Absolute Values Quiz

1) Find the limit as x approaches 0 from the right: lim_x→0⁺ (3x + 7)

0

3

7

10

The function 3x + 7 is a linear polynomial, which is continuous everywhere. To find the limit as x approaches 0 from the right, we substitute x = 0: 3(0) + 7 = 0 + 7 = 7. Therefore, the limit is 7.

Explanation

The function 3x + 7 is a linear polynomial, which is continuous everywhere. To find the limit as x approaches 0 from the right, we substitute x = 0: 3(0) + 7 = 0 + 7 = 7. Therefore, the limit is 7.

2) Evaluate: lim_x→(-2)⁺ (x + 2)/(|x + 2|)

-1

0

1

Does not exist

For x approaching -2 from the right (x > -2), x + 2 is positive. The absolute value of a positive number is the number itself, so |x + 2| = x + 2. Therefore, the expression becomes (x + 2)/(x + 2) = 1. So, the limit is 1.

Explanation

For x approaching -2 from the right (x > -2), x + 2 is positive. The absolute value of a positive number is the number itself, so |x + 2| = x + 2. Therefore, the expression becomes (x + 2)/(x + 2) = 1. So, the limit is 1.

3) Compute the left-hand limit: lim_x→1⁻ √(1 - x)

0

1

2

Does not exist

As x approaches 1 from the left, x is slightly less than 1, so 1 - x is a small positive number. The square root of a small positive number is also a small positive number approaching 0. Therefore, the limit is 0.

Explanation

As x approaches 1 from the left, x is slightly less than 1, so 1 - x is a small positive number. The square root of a small positive number is also a small positive number approaching 0. Therefore, the limit is 0.

4) A function's graph shows that as x approaches 2 from the left, the y-values get closer to 3. As x approaches 2 from the right, the y-values get closer to 3. What is lim_x→2⁻ f(x)?

2

3

6

Cannot be determined

The problem states that as x approaches 2 from the left, the y-values approach 3. This means the left-hand limit is 3.

Explanation

The problem states that as x approaches 2 from the left, the y-values approach 3. This means the left-hand limit is 3.

5) Which of the following must be true if the left-hand limit at x = c equals the right-hand limit at x = c?

The function is continuous at x = c.

The function is defined at x = c.

The overall limit at x = c exists.

The function is differentiable at x = c.

If the left-hand limit and the right-hand limit at x = c are equal to some number L, then the overall limit as x approaches c exists and equals L. The other options are not necessarily true because the function could still be undefined at c, or not continuous or differentiable even if the limit exists.

Explanation

If the left-hand limit and the right-hand limit at x = c are equal to some number L, then the overall limit as x approaches c exists and equals L. The other options are not necessarily true because the function could still be undefined at c, or not continuous or differentiable even if the limit exists.

6) The graph of a function has a vertical asymptote at x = 1. As x approaches 1 from the left, the graph goes downward indefinitely. What is lim_x→1⁻ f(x)?

1

0

-∞

∞

If the graph goes downward indefinitely as x approaches 1 from the left, this means the function values decrease without bound. Therefore, the left-hand limit is -∞.

Explanation

If the graph goes downward indefinitely as x approaches 1 from the left, this means the function values decrease without bound. Therefore, the left-hand limit is -∞.

7) Given f(x) = { 3x if x < 0; x² if x ≥ 0 }, find lim_x→0⁻ f(x).

0

1

3

Does not exist

The limit as x approaches 0 from the left uses the piece for x < 0, which is f(x) = 3x. Substitute x = 0: 3(0) = 0. Therefore, the left-hand limit is 0.

Explanation

The limit as x approaches 0 from the left uses the piece for x < 0, which is f(x) = 3x. Substitute x = 0: 3(0) = 0. Therefore, the left-hand limit is 0.

8) Evaluate: lim_x→4⁻ 1/(x - 4)

0

-∞

∞

1

As x approaches 4 from the left, x - 4 approaches 0 from the negative side (a very small negative number). Dividing 1 by a very small negative number yields a very large negative number. Therefore, the limit decreases without bound, which we denote as -∞.

Explanation

As x approaches 4 from the left, x - 4 approaches 0 from the negative side (a very small negative number). Dividing 1 by a very small negative number yields a very large negative number. Therefore, the limit decreases without bound, which we denote as -∞.

9) Evaluate lim_x→2⁻ (x + 2)/(x - 2).

-∞

+∞

4

Does not exist

As x approaches 2 from the left, numerator x + 2 → 4 (positive), denominator x - 2 is a small negative number. A positive divided by a small negative gives a large negative value, so the left-hand limit is -∞.

Explanation

As x approaches 2 from the left, numerator x + 2 → 4 (positive), denominator x - 2 is a small negative number. A positive divided by a small negative gives a large negative value, so the left-hand limit is -∞.

10) Consider f(x) = (|x + 3|)/(x + 3). Find lim_x→(-3)⁻ f(x).

-1

0

1

Does not exist

For x approaching -3 from the left (x < -3), x + 3 is negative. The absolute value of a negative number is its opposite, so |x + 3| = -(x + 3). Therefore, f(x) = [-(x + 3)]/(x + 3) = -1 for x < -3. So, as x approaches -3 from the left, f(x) approaches -1.

Explanation

For x approaching -3 from the left (x < -3), x + 3 is negative. The absolute value of a negative number is its opposite, so |x + 3| = -(x + 3). Therefore, f(x) = [-(x + 3)]/(x + 3) = -1 for x < -3. So, as x approaches -3 from the left, f(x) approaches -1.

11) Evaluate: lim_x→(-1)⁻(2x - 5)

-7

-5

-3

7

The function 2x - 5 is a linear polynomial, continuous everywhere. We substitute x = -1: 2(-1) - 5 = -2 - 5 = -7. Therefore, the limit as x approaches -1 from the left is -7.

Explanation

The function 2x - 5 is a linear polynomial, continuous everywhere. We substitute x = -1: 2(-1) - 5 = -2 - 5 = -7. Therefore, the limit as x approaches -1 from the left is -7.

12) Find the right-hand limit: lim_x→0⁺ (|x|)/x

-1

0

1

Does not exist

For x > 0 (approaching 0 from the right), |x| = x because x is positive. So, the expression becomes x/x = 1. Therefore, as x approaches 0 from the right, the function value is 1, so the limit is 1.

Explanation

For x > 0 (approaching 0 from the right), |x| = x because x is positive. So, the expression becomes x/x = 1. Therefore, as x approaches 0 from the right, the function value is 1, so the limit is 1.

13) Evaluate the limit: lim_x→2⁺ (x² - 4)/(x - 2)

0

2

4

Does not exist

Direct substitution gives 0/0. We factor the numerator: x² - 4 = (x - 2)(x + 2). The expression becomes (x - 2)(x + 2)/(x - 2). For x approaching 2 from the right (x > 2), x - 2 is not zero, so we cancel the common factor, leaving x + 2. Now take the limit: 2 + 2 = 4. Therefore, the limit is 4.

Explanation

Direct substitution gives 0/0. We factor the numerator: x² - 4 = (x - 2)(x + 2). The expression becomes (x - 2)(x + 2)/(x - 2). For x approaching 2 from the right (x > 2), x - 2 is not zero, so we cancel the common factor, leaving x + 2. Now take the limit: 2 + 2 = 4. Therefore, the limit is 4.

14) Compute: lim_x→5⁺ (10 - x)

5

10

15

0

The function 10 - x is a linear polynomial, continuous everywhere. Substitute x = 5: 10 - 5 = 5. Therefore, the limit as x approaches 5 from the right is 5.

Explanation

The function 10 - x is a linear polynomial, continuous everywhere. Substitute x = 5: 10 - 5 = 5. Therefore, the limit as x approaches 5 from the right is 5.

15) What is the relationship between one-sided limits and the value of the function at a point for continuity?

The one-sided limits must be equal to each other, but not necessarily equal to the function value.

The one-sided limits must be equal to the function value at that point.

The one-sided limits do not need to exist for the function to be continuous.

If the one-sided limits exist, the function must be continuous.

For a function to be continuous at a point x = c, three conditions must hold: f(c) is defined, the limit of f(x) as x approaches c exists, and the limit equals f(c). The limit exists if and only if the left-hand limit and right-hand limit exist and are equal. Therefore, both one-sided limits must equal the function value at that point for continuity.

Explanation

For a function to be continuous at a point x = c, three conditions must hold: f(c) is defined, the limit of f(x) as x approaches c exists, and the limit equals f(c). The limit exists if and only if the left-hand limit and right-hand limit exist and are equal. Therefore, both one-sided limits must equal the function value at that point for continuity.

One-Sided Limits with Absolute Values and Graph Behavior