One-Sided Limits Piecewise Functions Quiz

1) Consider the piecewise function f(x) = {x - 1 if x < 0, x + 1 if x >= 0}. What is lim x→0 f(x)?

Does not exist

-1

1

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 + 1, as x approaches 0 from the right, we plug in x=0 to get 0 + 1 = 1. Since -1 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→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 + 1, as x approaches 0 from the right, we plug in x=0 to get 0 + 1 = 1. Since -1 is not equal to 1, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

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

2

Does not exist

3

1

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, as x approaches 1 from the left, we plug in x=1 to get 2*1 = 2. 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 1² + 1 = 2. Since both the left-hand limit and the right-hand limit are 2, they are equal. Therefore, the overall limit is 2.

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, as x approaches 1 from the left, we plug in x=1 to get 2*1 = 2. 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 1² + 1 = 2. Since both the left-hand limit and the right-hand limit are 2, they are equal. Therefore, the overall limit is 2.

3) Given f(x) = {x³ if x < -1, -x if x >= -1}, find lim x→-1 f(x).

Does not exist

-1

1

0

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) = -x, as x approaches -1 from the right, we plug in x=-1 to get -(-1) = 1. Since -1 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→-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) = -x, as x approaches -1 from the right, we plug in x=-1 to get -(-1) = 1. Since -1 is not equal to 1, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

4) For the piecewise function f(x) = {3x + 2 if x ≤ 0, x if x > 0}, what is lim x→0 f(x)?

Does not exist

2

0

3

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

5) Consider f(x) = {x² if x < 4, 4x - 16 if x >= 4}. Find lim x→4 f(x).

Does not exist

16

0

4

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) = 4x - 16, as x approaches 4 from the right, we plug in x=4 to get 4*4 - 16 = 0. Since 16 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→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) = 4x - 16, as x approaches 4 from the right, we plug in x=4 to get 4*4 - 16 = 0. Since 16 is not equal to 0, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

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

5

Does not exist

2

3

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 = 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→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 = 5. Since both the left-hand limit and the right-hand limit are 5, they are equal. Therefore, the overall limit is 5.

7) Given f(x) = {1 if x < 3, x - 2 if x >= 3}, find lim x→3 f(x).

1

Does not exist

3

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 - 2, as x approaches 3 from the right, we plug in x=3 to get 3 - 2 = 1. Since both the left-hand limit and the right-hand limit are 1, they are equal. Therefore, the overall limit is 1.

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 - 2, as x approaches 3 from the right, we plug in x=3 to get 3 - 2 = 1. Since both the left-hand limit and the right-hand limit are 1, they are equal. Therefore, the overall limit is 1.

8) For f(x) = {sin(x) if x < 0, cos(x) if x >= 0}, what is lim x→0 f(x)?

Does not exist

0

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) = cos(x), as x approaches 0 from the right, we plug in x=0 to get cos(0) = 1. Since 0 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→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) = cos(x), as x approaches 0 from the right, we plug in x=0 to get cos(0) = 1. Since 0 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) = {2x² if x ≤ 1, 2 if x > 1}. Find lim x→1 f(x).

2

Does not exist

4

1

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) = 2x², as x approaches 1 from the left, we plug in x=1 to get 2*(1)² = 2. Next, the right-hand limit lim x→1+ f(x). Since for x > 1, f(x) = 2, as x approaches 1 from the right, f(x) approaches 2. Since both the left-hand limit and the right-hand limit are 2, they are equal. Therefore, the overall limit is 2.

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) = 2x², as x approaches 1 from the left, we plug in x=1 to get 2*(1)² = 2. Next, the right-hand limit lim x→1+ f(x). Since for x > 1, f(x) = 2, as x approaches 1 from the right, f(x) approaches 2. Since both the left-hand limit and the right-hand limit are 2, they are equal. Therefore, the overall limit is 2.

10) For f(x) = { -1 if x < 0, 1 if x > 0}, what is limx→0f(x)? (note: undefined at 0, but limit doesn't require it)

Does not exist

0

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) = 1, as x approaches 0 from the right, f(x) approaches 1. Since -1 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→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) = 1, as x approaches 0 from the right, f(x) approaches 1. Since -1 is not equal to 1, the left-hand limit and the right-hand limit are different. Therefore, the overall limit does not exist.

11) Consider f(x) = {x² - 1 if x < 1, x - 1 if x >= 1}. Find lim x→1 f(x).

0

Does not exist

1

-1

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) = x - 1, as x approaches 1 from the right, we plug in x=1 to get 1 - 1 = 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→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) = x - 1, as x approaches 1 from the right, we plug in x=1 to get 1 - 1 = 0. Since both the left-hand limit and the right-hand limit are 0, they are equal. Therefore, the overall limit is 0.

12) For a piecewise function, the limit at the switch point exists if which of the following is true?

The expressions from both sides approach the same value

The function value at the point matches one side

One side is constant and the other is linear

The limit is positive

For the limit at the switch point to exist, the left-hand limit from the left piece and the right-hand limit from the right piece must approach the same value. We calculate each one-sided limit by plugging the switch point into the respective expression. If they match, the limit exists and is that value; if not, it does not exist.

Explanation

For the limit at the switch point to exist, the left-hand limit from the left piece and the right-hand limit from the right piece must approach the same value. We calculate each one-sided limit by plugging the switch point into the respective expression. If they match, the limit exists and is that value; if not, it does not exist.

13) If the left-hand limit is L and the right-hand limit is M, and L = M, the limit exists even if the function is not defined at the point. Is this statement true or false?

True

False

Only if L is 0

Only if the function is defined

The statement is true because the limit exists if the left-hand limit equals the right-hand limit, say to L. The limit is then L, and this holds even if the function is not defined at the point, as the limit only concerns the approach from both sides, not the value at the point.

Explanation

The statement is true because the limit exists if the left-hand limit equals the right-hand limit, say to L. The limit is then L, and this holds even if the function is not defined at the point, as the limit only concerns the approach from both sides, not the value at the point.

14) What does it mean if the limit of a piecewise function at x=c exists?

The function approaches the same value from both sides

The left limit is greater than the right

The function value at c matches the limit

The function is constant

If the limit at x=c exists, it means the left-hand limit and the right-hand limit both approach the same value as x approaches c from the left and from the right. This is checked by evaluating the one-sided limits using the respective pieces, and if they agree, the limit exists.

Explanation

If the limit at x=c exists, it means the left-hand limit and the right-hand limit both approach the same value as x approaches c from the left and from the right. This is checked by evaluating the one-sided limits using the respective pieces, and if they agree, the limit exists.

15) In piecewise functions, the limit at the boundary point can exist even if the two pieces are different functions, as long as what?

They approach the same value at the boundary

They are both polynomials

The boundary value is zero

The functions are the same

The limit at the boundary point exists if the left piece approaches a value and the right piece approaches the same value as x approaches the boundary from each side. The types of functions in the pieces do not matter; we just evaluate the one-sided limits, and if they are equal, the limit exists.

Explanation

The limit at the boundary point exists if the left piece approaches a value and the right piece approaches the same value as x approaches the boundary from each side. The types of functions in the pieces do not matter; we just evaluate the one-sided limits, and if they are equal, the limit exists.

One-Sided Limits in Piecewise Functions