One-Sided Limits Algebraic Functions Quiz

1) Find lim_x→3⁻ (5x - 2)

13

15

17

Does not exist

To find the limit as x approaches 3 from the left, we substitute x = 3 into the expression because the function 5x - 2 is a polynomial, which is continuous everywhere. Substituting gives 5(3) - 2 = 15 - 2 = 13. Therefore, the limit is 13.

Explanation

To find the limit as x approaches 3 from the left, we substitute x = 3 into the expression because the function 5x - 2 is a polynomial, which is continuous everywhere. Substituting gives 5(3) - 2 = 15 - 2 = 13. Therefore, the limit is 13.

2) Evaluate the one-sided limit: lim_x→0⁺ (1/x)

0

1

∞

-∞

As x approaches 0 from the positive side (right), x becomes a very small positive number. Dividing 1 by a very small positive number yields a very large positive number. Therefore, the limit increases without bound, which we denote as ∞.

Explanation

As x approaches 0 from the positive side (right), x becomes a very small positive number. Dividing 1 by a very small positive number yields a very large positive number. Therefore, the limit increases without bound, which we denote as ∞.

3) Evaluate lim_x→5-(x - 5)/|x - 5|.

-1

1

0

Does not exist

For x < 5, x - 5 is negative, so |x - 5| = -(x - 5). The ratio becomes (x - 5)/ (-(x - 5)) = -1 for x < 5. Approaching 5 from the left the expression remains -1, so the left-hand limit equals -1.

Explanation

For x < 5, x - 5 is negative, so |x - 5| = -(x - 5). The ratio becomes (x - 5)/ (-(x - 5)) = -1 for x < 5. Approaching 5 from the left the expression remains -1, so the left-hand limit equals -1.

4) Determine the right-hand limit: lim_x→4⁺√(x - 4)

0

2

4

Does not exist

As x approaches 4 from the right, x - 4 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 4 from the right, x - 4 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.

5) Consider the function f(x) = (|x - 5|)/(x - 5). Find lim_x→5⁺ f(x).

-1

0

1

Does not exist

For x > 5 (approaching 5 from the right), x - 5 is positive. The absolute value of a positive number is the number itself, so |x - 5| = x - 5. Therefore, f(x) = (x - 5)/(x - 5) = 1 for x > 5. So, as x approaches 5 from the right, f(x) approaches 1.

Explanation

For x > 5 (approaching 5 from the right), x - 5 is positive. The absolute value of a positive number is the number itself, so |x - 5| = x - 5. Therefore, f(x) = (x - 5)/(x - 5) = 1 for x > 5. So, as x approaches 5 from the right, f(x) approaches 1.

6) A function g(x) has a graph that approaches the value 2 as x approaches 1 from the left, but approaches the value 5 as x approaches 1 from the right. What is lim_x→1⁻ g(x)?

1

2

5

Does not exist

The limit of g(x) as x approaches 1 from the left is the value that the graph approaches from the left side. The problem states that as x approaches 1 from the left, the graph approaches 2. Therefore, the left-hand limit is 2.

Explanation

The limit of g(x) as x approaches 1 from the left is the value that the graph approaches from the left side. The problem states that as x approaches 1 from the left, the graph approaches 2. Therefore, the left-hand limit is 2.

7) Evaluate: lim_x→(-2)⁻ (x² + 1)

3

5

6

7

The function x² + 1 is a polynomial, which is continuous everywhere. To find the limit as x approaches -2 from the left, we can directly substitute x = -2. Calculation: (-2)² + 1 = 4 + 1 = 5. Therefore, the limit is 5.

Explanation

The function x² + 1 is a polynomial, which is continuous everywhere. To find the limit as x approaches -2 from the left, we can directly substitute x = -2. Calculation: (-2)² + 1 = 4 + 1 = 5. Therefore, the limit is 5.

8) Find the left-hand limit: lim_x→0⁻ (|x|)/x

-1

0

1

Does not exist

For x < 0 (approaching 0 from the left), |x| = -x because x is negative. So, the expression becomes (-x)/x = -1. Therefore, as x approaches 0 from the left, the function value is -1, so the limit is -1.

Explanation

For x < 0 (approaching 0 from the left), |x| = -x because x is negative. So, the expression becomes (-x)/x = -1. Therefore, as x approaches 0 from the left, the function value is -1, so the limit is -1.

9) If a function f(x) is continuous at x = a, what must be true about the one-sided limits at x = a?

Limx→a⁻ f(x) must equal limx→a⁺ f(x)

Limx→a⁻ f(x) must be greater than limx→a⁺ f(x)

Limx→a⁻ f(x) must be less than limx→a⁺ f(x)

The one-sided limits do not need to be equal.

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

Explanation

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

10) Evaluate the limit: lim_x→1⁺ (x² - 1)/(x - 1)

0

1

2

Does not exist

First, note that direct substitution gives 0/0, which is indeterminate. We can factor the numerator: x² - 1 = (x - 1)(x + 1). So, the expression becomes (x - 1)(x + 1)/(x - 1). For x approaching 1 from the right (x > 1), x - 1 is not zero, so we can cancel the common factor. This simplifies to x + 1. Now, take the limit as x approaches 1: 1 + 1 = 2. Therefore, the limit is 2.

Explanation

First, note that direct substitution gives 0/0, which is indeterminate. We can factor the numerator: x² - 1 = (x - 1)(x + 1). So, the expression becomes (x - 1)(x + 1)/(x - 1). For x approaching 1 from the right (x > 1), x - 1 is not zero, so we can cancel the common factor. This simplifies to x + 1. Now, take the limit as x approaches 1: 1 + 1 = 2. Therefore, the limit is 2.

11) The graph of function h(x) shows a hole at the point (3, 4) and the curve approaches this hole from both sides. What is lim_x→3⁻ h(x)?

3

4

0

Does not exist

A hole in the graph at (3, 4) indicates that the function is not defined at x = 3, but the limit as x approaches 3 exists and equals the y-coordinate of the hole, which is 4. Since the curve approaches this hole from both the left and the right, the left-hand limit as x approaches 3 is also 4.

Explanation

A hole in the graph at (3, 4) indicates that the function is not defined at x = 3, but the limit as x approaches 3 exists and equals the y-coordinate of the hole, which is 4. Since the curve approaches this hole from both the left and the right, the left-hand limit as x approaches 3 is also 4.

12) Compute: lim_x→(-3)⁺(x + 4)

-1

0

1

7

The function x + 4 is a linear polynomial, which is continuous everywhere. To find the limit as x approaches -3 from the right, we substitute x = -3: (-3) + 4 = 1. Therefore, the limit is 1.

Explanation

The function x + 4 is a linear polynomial, which is continuous everywhere. To find the limit as x approaches -3 from the right, we substitute x = -3: (-3) + 4 = 1. Therefore, the limit is 1.

13) For the function f(x) = { 2x if x ≤ 1; x² + 1 if x > 1 }, find lim_x→1⁺ f(x).

1

2

3

4

The limit as x approaches 1 from the right uses the piece of the function defined for x > 1, which is f(x) = x² + 1. We substitute x = 1 into this expression: (1)² + 1 = 1 + 1 = 2. Therefore, the right-hand limit is 2.

Explanation

The limit as x approaches 1 from the right uses the piece of the function defined for x > 1, which is f(x) = x² + 1. We substitute x = 1 into this expression: (1)² + 1 = 1 + 1 = 2. Therefore, the right-hand limit is 2.

14) If the left-hand limit and the right-hand limit at x = c are not equal, what can be said about the overall limit at x = c?

The overall limit equals the left-hand limit.

The overall limit equals the right-hand limit.

The overall limit equals the average of the two one-sided limits.

The overall limit does not exist.

For the overall limit to exist at a point, the left-hand limit and the right-hand limit must both exist and be equal to each other. If they are not equal, then the overall limit does not exist.

Explanation

For the overall limit to exist at a point, the left-hand limit and the right-hand limit must both exist and be equal to each other. If they are not equal, then the overall limit does not exist.

15) Evaluate: lim_x→2⁻ (x - 2)/(|x - 2|)

-1

0

1

Does not exist

For x approaching 2 from the left (x < 2), x - 2 is negative. The absolute value of a negative number is its opposite, 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 left (x < 2), x - 2 is negative. The absolute value of a negative number is its opposite, so |x - 2| = -(x - 2). Therefore, the expression becomes (x - 2)/[-(x - 2)] = -1. So, the limit is -1.

One-Sided Limits with Basic Algebraic Functions