One-Sided Limits Holes & Infinite Behavior Quiz

1) Find lim_x→1⁻ (4x + 3)

1

4

7

12

The function 4x + 3 is a linear polynomial, continuous everywhere. Substitute x = 1: 4(1) + 3 = 4 + 3 = 7. Therefore, the limit as x approaches 1 from the left is 7.

Explanation

The function 4x + 3 is a linear polynomial, continuous everywhere. Substitute x = 1: 4(1) + 3 = 4 + 3 = 7. Therefore, the limit as x approaches 1 from the left is 7.

2) Evaluate: lim_x→0⁻ 1/x

0

1

∞

-∞

As x approaches 0 from the left, x is 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, denoted as -∞.

Explanation

As x approaches 0 from the left, x is 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, denoted as -∞.

3) Compute the right-hand limit: lim_x→9⁺ √(x - 9)

0

3

9

Does not exist

As x approaches 9 from the right, x - 9 approaches 0 from the positive side. The square root of a number approaching 0 from the positive side is also a number approaching 0. Therefore, the limit is 0.

Explanation

As x approaches 9 from the right, x - 9 approaches 0 from the positive side. The square root of a number approaching 0 from the positive side is also a number approaching 0. Therefore, the limit is 0.

4) Consider f(x) = (|x - 1|)/(x - 1). Find lim_x→1⁺ f(x).

-1

0

1

Does not exist

For x > 1 (approaching 1 from the right), x - 1 is positive. The absolute value of a positive number is the number itself, so |x - 1| = x - 1. Therefore, f(x) = (x - 1)/(x - 1) = 1 for x > 1. So, as x approaches 1 from the right, f(x) approaches 1.

Explanation

For x > 1 (approaching 1 from the right), x - 1 is positive. The absolute value of a positive number is the number itself, so |x - 1| = x - 1. Therefore, f(x) = (x - 1)/(x - 1) = 1 for x > 1. So, as x approaches 1 from the right, f(x) approaches 1.

5) A function h(x) has a graph that approaches the value -2 as x approaches 0 from the left and approaches the value 4 as x approaches 0 from the right. What is lim_x→0⁺ h(x)?

-2

0

2

4

The limit as x approaches 0 from the right is the value that the graph approaches from the right side. The problem states that as x approaches 0 from the right, the graph approaches 4. Therefore, the right-hand limit is 4.

Explanation

The limit as x approaches 0 from the right is the value that the graph approaches from the right side. The problem states that as x approaches 0 from the right, the graph approaches 4. Therefore, the right-hand limit is 4.

6) Evaluate: lim_x→(-5)⁻ (x² + 2x)

15

20

25

35

The function x² + 2x is a polynomial, continuous everywhere. Substitute x = -5: (-5)² + 2(-5) = 25 - 10 = 15. Therefore, the limit as x approaches -5 from the left is 15.

Explanation

The function x² + 2x is a polynomial, continuous everywhere. Substitute x = -5: (-5)² + 2(-5) = 25 - 10 = 15. Therefore, the limit as x approaches -5 from the left is 15.

7) Find the left-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, 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, the limit is 1.

8) Evaluate lim_x→0- 1/|x|.

0

+∞

-∞

1

For x approaching 0 from the left, x is negative, but |x| = -x is positive and small. The reciprocal 1/|x| becomes a very large positive number. Therefore the left-hand limit is +∞.

Explanation

For x approaching 0 from the left, x is negative, but |x| = -x is positive and small. The reciprocal 1/|x| becomes a very large positive number. Therefore the left-hand limit is +∞.

9) If the overall limit at x = a exists, what must be true about the one-sided limits?

The left-hand limit must be greater than the right-hand limit.

The left-hand limit must be less than the right-hand limit.

The left-hand limit and right-hand limit must be equal.

At least one one-sided limit must exist.

The overall limit at x = a exists if and only if both one-sided limits exist and are equal to each other. Therefore, they must be equal.

Explanation

The overall limit at x = a exists if and only if both one-sided limits exist and are equal to each other. Therefore, they must be equal.

10) Evaluate lim_x→(-2)⁺(x² - 4)/(x + 2)

-4

-2

0

4

Direct substitution gives 0/0. 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.

Explanation

Direct substitution gives 0/0. 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.

11) The graph of a function has a hole at (2, 5). What is lim_x→2⁻ f(x)?

2

5

0

Does not exist

A hole at (2, 5) indicates that the function is not defined at x = 2, but the limit as x approaches 2 exists and equals the y-coordinate of the hole. Since the hole is at y = 5, and assuming the graph approaches this hole from both sides, the left-hand limit is also 5.

Explanation

A hole at (2, 5) indicates that the function is not defined at x = 2, but the limit as x approaches 2 exists and equals the y-coordinate of the hole. Since the hole is at y = 5, and assuming the graph approaches this hole from both sides, the left-hand limit is also 5.

12) Compute: lim_x→6⁻ (x - 3)

3

6

9

0

The function x - 3 is a linear polynomial, continuous everywhere. Substitute x = 6: 6 - 3 = 3. Therefore, the limit as x approaches 6 from the left is 3.

Explanation

The function x - 3 is a linear polynomial, continuous everywhere. Substitute x = 6: 6 - 3 = 3. Therefore, the limit as x approaches 6 from the left is 3.

13) Given f(x) = { x³ if x ≤ 0; 2x if x > 0 }, find lim_x→0⁺ f(x).

0

1

2

Does not exist

The limit as x approaches 0 from the right uses the piece for x > 0, which is f(x) = 2x. Substitute x = 0: 2(0) = 0. Therefore, the right-hand limit is 0.

Explanation

The limit as x approaches 0 from the right uses the piece for x > 0, which is f(x) = 2x. Substitute x = 0: 2(0) = 0. Therefore, the right-hand limit is 0.

14) Which statement best describes the concept of a one-sided limit?

It is the value that a function approaches as the input approaches a specific value from one side only.

It is the actual value of the function at a specific point.

It is always equal to the overall limit.

It only exists for continuous functions.

A one-sided limit describes the behavior of a function as the input approaches a particular point from either the left or the right. It does not have to equal the function value at that point, and it can exist even if the overall limit does not exist.

Explanation

A one-sided limit describes the behavior of a function as the input approaches a particular point from either the left or the right. It does not have to equal the function value at that point, and it can exist even if the overall limit does not exist.

15) Evaluate: lim_x→3⁻ (x - 3)/(|x - 3|)

-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, the expression becomes (x - 3)/[-(x - 3)] = -1. So, the limit is -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, the expression becomes (x - 3)/[-(x - 3)] = -1. So, the limit is -1.

One-Sided Limits Involving Holes and Infinite Behavior