Matching Left & Right Limits Quiz

1) Consider the piecewise function f(x) = {x + 4 if x < -3, x² if x >= -3}. What is lim x→-3 f(x)?

Does not exist

1

9

-3

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 f(x) = {3x if x ≤ 2, 6 if x > 2}, what is lim x→2 f(x)?

6

Does not exist

3

2

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) Given f(x) = {x - 5 if x < 0, -x - 5 if x >= 0}, find lim x→0 f(x).

-5

Does not exist

0

5

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) For the piecewise function f(x) = {4 if x < 1, 2x + 2 if x >= 1}, what is lim x→1 f(x)?

4

Does not exist

1

6

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) Consider f(x) = {x³ if x < 3, x if x >= 3}. Find lim x→3 f(x).

Does not exist

27

3

0

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) For f(x) = {2x - 1 if x ≤ -1, x² - 1 if x > -1}, what is lim x→-1 f(x)?

Does not exist

-3

0

-2

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) Given f(x) = {x + 1 if x < 4, 5 if x >= 4}, find lim x→4 f(x).

5

Does not exist

4

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) = {1/x if x < 2, x - 3 if x >= 2}, what is lim x→2 f(x)?

Does not exist

0.5

-1

∞

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) Consider f(x) = {x² if x ≤ 0, -x² if x > 0}. Find lim x→0 f(x).

0

Does not exist

1

-1

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) For f(x) = {4x if x < 3, x + 9 if x >= 3}, what is lim x→3 f(x)?

12

Does not exist

3

6

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) Consider f(x) = {x - 2 if x < 0, -x + 2 if x >= 0}. Find lim x→0 f(x).

Does not exist

-2

0

2

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) Why do we need to check both left and right limits for piecewise functions?

Because the function may approach different values from each side

Because the function is always continuous

Because limits only exist from one side

Because the point value matters for the limit

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) If the one-sided limits agree but differ from f(c), the limit still exists. Is this statement true or false?

True

False

Only for polynomials

Only if zero

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 piecewise functions, the limit does not exist when?

The one-sided limits differ

The function is defined at the point

Both pieces are the same

The limit is finite

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) A piecewise function's limit at the join point exists if the values from both pieces match at the approach, regardless of function types. Is this statement true or false?

True

False

Only for linear functions

Only if continuous

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.