Topic: Jump discontinuities
Question: Which of the following statements is true?
Answer choices:
A A jump discontinuity occurs when the right-hand limit and left-hand
limit of a function are different but they both exist.
B A jump discontinuity occurs when the right-hand limit and left-hand
limit of a function are different but only one exists.
C A jump discontinuity occurs when the right-hand limit and left-hand
limit of a function do not exist.
D A jump discontinuity occurs when the right-hand limit and left-hand
limit of a function are the same and they both exist.
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Solution: A
A jump discontinuity occurs when the right-hand limit and left-hand limit of
a function both exist, but have different values.
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Topic: Jump discontinuities
Question: Choose the correct description of the jump discontinuity.
x
f (x) =
|x|
Answer choices:
A The function has a jump discontinuity at x = 1.
B The function has a jump discontinuity at x = − 1.
C The function has a jump discontinuity at x = ∞.
D The function has a jump discontinuity at x = 0.
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Solution: D
To solve for the discontinuities of the function, start by evaluating the
right-hand and left-hand limits as x approaches 0. We need to use 0 for the
limit because the function does not exist when x = 0.
For the right-hand limit, we’ll use 3, 2 and 1.
x 3 3
lim+ = = =1
x→0 |x| |3| 3
x 2 2
lim+ = = =1
x→0 |x| |2| 2
x 1 1
lim+ = = =1
x→0 |x| |1| 1
For the left-hand limit, we’ll use −3, −2 and −1.
x −3 −3
lim− = = =−1
x→0 |x| | − 3| 3
x −2 −2
lim = = =−1
x→0− | x | | − 2| 2
x −1 −1
lim = = =−1
x→0− | x | | − 1| 1
We can see that both one-sided limits exist. The right-hand limit is 1 and
the left-hand limit is −1. Because both limits exist but aren’t equal to each
other, there’s a jump discontinuity at 0.
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Topic: Jump discontinuities
Question: Choose the correct description of the jump discontinuity.
x−1
f (x) =
|x − 1|
Answer choices:
A The function has a jump discontinuity at x = − 1.
B The function has a jump discontinuity at x = 1.
C The function has a jump discontinuity at x = ∞.
D The function has a jump discontinuity at x = 0.
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Solution: B
To solve for the discontinuities of the function, start by evaluating the
right-hand and left-hand limits as x approaches 1. We need to use 1 for the
limit because the function does not exist when x = 1.
For the right-hand limit, we’ll use 4, 3 and 2.
x−1 (4) − 1 3 3
lim+ = = = =1
x→1 |x − 1| | (4) − 1 | |3| 3
x−1 (3) − 1 2 2
lim+ = = = =1
x→1 |x − 1| | (3) − 1 | |2| 2
x−1 (2) − 1 1 1
lim+ = = = =1
x→1 |x − 1| | (2) − 1 | |1| 1
For the left-hand limit, we’ll use −2, −1 and 0.
x−1 (−2) − 1 −3 −3
lim− = = = =−1
x→1 |x − 1| | (−2) − 1 | | − 3| 3
x−1 (−1) − 1 −2 −2
lim = = = =−1
x→1− | x − 1 | | (−1) − 1 | | − 2| 2
x−1 (0) − 1 −1 −1
lim = = = =−1
x→1− | x − 1 | | (0) − 1 | | − 1| 1
We can see that both one-sided limits exist. The right-hand limit is 1 and
the left-hand limit is −1. Because both limits exist but aren’t equal to each
other, there’s a jump discontinuity at −1.
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