Advanced Piecewise Limits Quiz

1) Consider g(x) = {2 - x² if x < -1, x³ if x ≥ -1}. Find the limit as x approaches -1.

1

-1

0

Does not exist

To find lim x→-3 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→-3- f(x). Since for x = -3, f(x) = x², as x approaches -3 from the right, we plug in x=-3 to get (-3)² = 9. Since 1 is not equal to 9, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

Explanation

To find lim x→-3 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→-3- f(x). Since for x = -3, f(x) = x², as x approaches -3 from the right, we plug in x=-3 to get (-3)² = 9. Since 1 is not equal to 9, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

2) For h(x) = {4x if x < 3, x² + 3 if x ≥ 3}, what is the limit as x approaches 3?

12

Does not exist

9

6

To find lim x→2 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→2- f(x). Since for x ≤ 2, f(x) = 3x, as x approaches 2 from the left, we plug in x=2 to get 3*2 = 6. Next, the right-hand limit lim x→2+ f(x). Since for x > 2, f(x) = 6, as x approaches 2 from the right, f(x) approaches 6. Since both the left-hand limit and the right-hand limit are 6, they are equal. Therefore, the overall limit is 6.

Explanation

To find lim x→2 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→2- f(x). Since for x ≤ 2, f(x) = 3x, as x approaches 2 from the left, we plug in x=2 to get 3*2 = 6. Next, the right-hand limit lim x→2+ f(x). Since for x > 2, f(x) = 6, as x approaches 2 from the right, f(x) approaches 6. Since both the left-hand limit and the right-hand limit are 6, they are equal. Therefore, the overall limit is 6.

3) Let f(x) = {(x² - 16)/(x - 4) if x ≠ 4, 7 if x = 4}. Find the limit as x approaches 4.

7

4

8

0

To find lim x→0 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→0- f(x). Since for x = 0, f(x) = -x - 5, as x approaches 0 from the right, we plug in x=0 to get -0 - 5 = -5. Since both the left-hand limit and the right-hand limit are -5, they are equal. Therefore, the overall limit is -5.

Explanation

To find lim x→0 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→0- f(x). Since for x = 0, f(x) = -x - 5, as x approaches 0 from the right, we plug in x=0 to get -0 - 5 = -5. Since both the left-hand limit and the right-hand limit are -5, they are equal. Therefore, the overall limit is -5.

4) Consider g(x) = {|x + 2|/(x + 2) if x ≠ -2, 0 if x = -2}. What is the limit as x approaches -2?

0

1

-1

Does not exist

To find lim x→1 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→1- f(x). Since for x = 1, f(x) = 2x + 2, as x approaches 1 from the right, we plug in x=1 to get 2*1 + 2 = 4. Since both the left-hand limit and the right-hand limit are 4, they are equal. Therefore, the overall limit is 4.

Explanation

To find lim x→1 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→1- f(x). Since for x = 1, f(x) = 2x + 2, as x approaches 1 from the right, we plug in x=1 to get 2*1 + 2 = 4. Since both the left-hand limit and the right-hand limit are 4, they are equal. Therefore, the overall limit is 4.

5) For f(x) = {x² - 2x if x ≤ 2, 2x if x > 2}, evaluate the limit as x approaches 2.

0

4

Does not exist

2

To find lim x→3 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→3- f(x). Since for x = 3, f(x) = x, as x approaches 3 from the right, we plug in x=3 to get 3. Since 27 is not equal to 3, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

Explanation

To find lim x→3 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→3- f(x). Since for x = 3, f(x) = x, as x approaches 3 from the right, we plug in x=3 to get 3. Since 27 is not equal to 3, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

6) Let h(x) = {(x² - 25)/(x - 5) if x < 5, 3x - 5 if x ≥ 5}. Find the limit as x approaches 5.

10

Does not exist

15

5

To find lim x→-1 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→-1- f(x). Since for x ≤ -1, f(x) = 2x - 1, as x approaches -1 from the left, we plug in x=-1 to get 2*(-1) - 1 = -3. Next, the right-hand limit lim x→-1+ f(x). Since for x > -1, f(x) = x² - 1, as x approaches -1 from the right, we plug in x=-1 to get (-1)² - 1 = 0. Since -3 is not equal to 0, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

Explanation

To find lim x→-1 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→-1- f(x). Since for x ≤ -1, f(x) = 2x - 1, as x approaches -1 from the left, we plug in x=-1 to get 2*(-1) - 1 = -3. Next, the right-hand limit lim x→-1+ f(x). Since for x > -1, f(x) = x² - 1, as x approaches -1 from the right, we plug in x=-1 to get (-1)² - 1 = 0. Since -3 is not equal to 0, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

7) Consider g(x) = {√(9 - x²) if -3 ≤ x ≤ 3, x + 6 if x < -3 or x > 3}. What is the limit as x approaches 3 from the left?

3

0

9

6

To find lim x→4 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→4- f(x). Since for x = 4, f(x) = 5, as x approaches 4 from the right, f(x) approaches 5. Since both the left-hand limit and the right-hand limit are 5, they are equal. Therefore, the overall limit is 5.

Explanation

To find lim x→4 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→4- f(x). Since for x = 4, f(x) = 5, as x approaches 4 from the right, f(x) approaches 5. Since both the left-hand limit and the right-hand limit are 5, they are equal. Therefore, the overall limit is 5.

8) For f(x) = {5 if x < 0, x² + 5 if 0 ≤ x < 3, 3x - 1 if x ≥ 3}, find the limit as x approaches 0.

5

0

Does not exist

4

To find lim x→2 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→2- f(x). Since for x = 2, f(x) = x - 3, as x approaches 2 from the right, we plug in x=2 to get 2 - 3 = -1. Since 0.5 is not equal to -1, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

Explanation

To find lim x→2 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→2- f(x). Since for x = 2, f(x) = x - 3, as x approaches 2 from the right, we plug in x=2 to get 2 - 3 = -1. Since 0.5 is not equal to -1, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

9) Let h(x) = {cos(πx)/x if x ≠ 0, 1 if x = 0}. What is the limit as x approaches 0?

π

0

1

Does not exist

To find lim x→0 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→0- f(x). Since for x ≤ 0, f(x) = x², as x approaches 0 from the left, we plug in x=0 to get 0² = 0. Next, the right-hand limit lim x→0+ f(x). Since for x > 0, f(x) = -x², as x approaches 0 from the right, we plug in x=0 to get - (0)² = 0. Since both the left-hand limit and the right-hand limit are 0, they are equal. Therefore, the overall limit is 0.

Explanation

To find lim x→0 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→0- f(x). Since for x ≤ 0, f(x) = x², as x approaches 0 from the left, we plug in x=0 to get 0² = 0. Next, the right-hand limit lim x→0+ f(x). Since for x > 0, f(x) = -x², as x approaches 0 from the right, we plug in x=0 to get - (0)² = 0. Since both the left-hand limit and the right-hand limit are 0, they are equal. Therefore, the overall limit is 0.

10) Consider f(x) = {(x³ - 27)/(x - 3) if x ≠ 3, 25 if x = 3}. Find the limit as x approaches 3.

25

9

27

18

To find lim x→3 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→3- f(x). Since for x = 3, f(x) = x + 9, as x approaches 3 from the right, we plug in x=3 to get 3 + 9 = 12. Since both the left-hand limit and the right-hand limit are 12, they are equal. Therefore, the overall limit is 12.

Explanation

To find lim x→3 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→3- f(x). Since for x = 3, f(x) = x + 9, as x approaches 3 from the right, we plug in x=3 to get 3 + 9 = 12. Since both the left-hand limit and the right-hand limit are 12, they are equal. Therefore, the overall limit is 12.

11) For g(x) = {(√(x + 5) - 3)/(x - 4) if x > 0 and x ≠ 4, 1/6 if x = 4}, what is the limit as x approaches 4?

1/6

1/3

1/12

0

To find lim x→0 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→0- f(x). Since for x = 0, f(x) = -x + 2, as x approaches 0 from the right, we plug in x=0 to get -0 + 2 = 2. Since -2 is not equal to 2, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

Explanation

To find lim x→0 f(x), we evaluate the left-hand limit and the right-hand limit. We start with the left-hand limit lim x→0- f(x). Since for x = 0, f(x) = -x + 2, as x approaches 0 from the right, we plug in x=0 to get -0 + 2 = 2. Since -2 is not equal to 2, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

12) Let f(x) = {|x + 1| if x ≠ -1, 2 if x = -1}. Evaluate the limit as x approaches -1.

2

-1

0

1

For piecewise functions, the expression changes at a point, so the left-hand limit using the left piece may differ from the right-hand limit using the right piece. To see if the limit exists, we calculate both and check if they are equal. If they are, the limit exists; if not, it does not, as the function approaches different values from each side.

Explanation

For piecewise functions, the expression changes at a point, so the left-hand limit using the left piece may differ from the right-hand limit using the right piece. To see if the limit exists, we calculate both and check if they are equal. If they are, the limit exists; if not, it does not, as the function approaches different values from each side.

13) Consider h(x) = {(x² - x)/(x - 1) if x < 1, bx² if x ≥ 1}. For what value of b does the limit as x approaches 1 exist?

B = 0

B = 1

B = 2

B = 1/2

The statement is true because if the left-hand limit and the right-hand limit agree on a value, the overall limit exists and is that value. This is independent of f(c), as the limit is determined only by the approach from both sides, not the function's value at c.

Explanation

The statement is true because if the left-hand limit and the right-hand limit agree on a value, the overall limit exists and is that value. This is independent of f(c), as the limit is determined only by the approach from both sides, not the function's value at c.

14) For f(x) = {(3x² + 2x - 1)/(x + 1) if x ≠ -1, m if x = -1}, what value of m makes f continuous at x = -1?

M = -4

M = -2

M = 0

M = 2

The limit does not exist when the left-hand limit and the right-hand limit are not the same. We find this by evaluating each using the appropriate piece, and if the values differ, the limit does not exist because the function does not approach a single value.

Explanation

The limit does not exist when the left-hand limit and the right-hand limit are not the same. We find this by evaluating each using the appropriate piece, and if the values differ, the limit does not exist because the function does not approach a single value.

15) Let g(x) = {(x² - 4x + 4)/(x - 2) if x < 2, ax + 2 if x ≥ 2}. For what value of a does the limit as x approaches 2 exist?

A = -1

A = 0

A = 1

A = 2

The statement is true because the limit exists if the left-hand limit from one piece equals the right-hand limit from the other piece, even if the pieces are different types of functions. We evaluate each one-sided limit by substituting the join point into the respective expression, and if they match, the limit exists. The types of functions do not affect this condition.

Explanation

The statement is true because the limit exists if the left-hand limit from one piece equals the right-hand limit from the other piece, even if the pieces are different types of functions. We evaluate each one-sided limit by substituting the join point into the respective expression, and if they match, the limit exists. The types of functions do not affect this condition.

Advanced Piecewise Limit Challenges

1) Consider g(x) = {2 - x² if x < -1, x³ if x ≥ -1}. Find the limit as x approaches -1.

2)

What first name or nickname would you like us to use?

2) For h(x) = {4x if x < 3, x² + 3 if x ≥ 3}, what is the limit as x approaches 3?

3) Let f(x) = {(x² - 16)/(x - 4) if x ≠ 4, 7 if x = 4}. Find the limit as x approaches 4.

4) Consider g(x) = {|x + 2|/(x + 2) if x ≠ -2, 0 if x = -2}. What is the limit as x approaches -2?

5) For f(x) = {x² - 2x if x ≤ 2, 2x if x > 2}, evaluate the limit as x approaches 2.

6) Let h(x) = {(x² - 25)/(x - 5) if x < 5, 3x - 5 if x ≥ 5}. Find the limit as x approaches 5.

7) Consider g(x) = {√(9 - x²) if -3 ≤ x ≤ 3, x + 6 if x < -3 or x > 3}. What is the limit as x approaches 3 from the left?

8) For f(x) = {5 if x < 0, x² + 5 if 0 ≤ x < 3, 3x - 1 if x ≥ 3}, find the limit as x approaches 0.

9) Let h(x) = {cos(πx)/x if x ≠ 0, 1 if x = 0}. What is the limit as x approaches 0?

10) Consider f(x) = {(x³ - 27)/(x - 3) if x ≠ 3, 25 if x = 3}. Find the limit as x approaches 3.

11) For g(x) = {(√(x + 5) - 3)/(x - 4) if x > 0 and x ≠ 4, 1/6 if x = 4}, what is the limit as x approaches 4?

12) Let f(x) = {|x + 1| if x ≠ -1, 2 if x = -1}. Evaluate the limit as x approaches -1.

13) Consider h(x) = {(x² - x)/(x - 1) if x < 1, bx² if x ≥ 1}. For what value of b does the limit as x approaches 1 exist?

14) For f(x) = {(3x² + 2x - 1)/(x + 1) if x ≠ -1, m if x = -1}, what value of m makes f continuous at x = -1?

15) Let g(x) = {(x² - 4x + 4)/(x - 2) if x < 2, ax + 2 if x ≥ 2}. For what value of a does the limit as x approaches 2 exist?