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Exercises for Limit Laws-1
Exercises for Limit LawsFind the indicated limits:
(1) limx→1
[x5 − 3x3 + 1
](x2 − 2) Solution
(2) limx→16
√x
x + 16Solution
(3) limx→16
√x − 4x − 16
Solution
(4) limx→3
x − 3x2 − 9
Solution
(5) limx→2
x2 − 4x − 4
= Solution
Exercises for Limit Laws-2
(6) limx→1−
⌊x − 12
⌋Solution
(7) limx→1−
|x − 1|x − 1
Solution
(8) limx→1+
|x − 1|x − 1
Solution
(9) limx→∞
x2 + 1+ 11− x Solution
(10) limx→−∞
x − 1x2 − 3x − 1
Solution
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Exercises for Limit Laws-3
Solutions
(1)limx→1
[x5 − 3x3 + 1
](x2 − 2) =
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Exercises for Limit Laws-3
Solutions
(1)limx→1
[x5 − 3x3 + 1
](x2 − 2) =[(1)5 − 3(1)3 + 1
]((1)2 − 2) =
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Exercises for Limit Laws-3
Solutions
(1)limx→1
[x5 − 3x3 + 1
](x2 − 2) =[(1)5 − 3(1)3 + 1
]((1)2 − 2) =
[1− 3+ 1] (1− 2)
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Exercises for Limit Laws-3
Solutions
(1)limx→1
[x5 − 3x3 + 1
](x2 − 2) =[(1)5 − 3(1)3 + 1
]((1)2 − 2) =
[1− 3+ 1] (1− 2) = [−1] (−1) =
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Exercises for Limit Laws-3
Solutions
(1)limx→1
[x5 − 3x3 + 1
](x2 − 2) =[(1)5 − 3(1)3 + 1
]((1)2 − 2) =
[1− 3+ 1] (1− 2) = [−1] (−1) = 1
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Exercises for Limit Laws-4
(2)limx→16
√x
x + 16=
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Exercises for Limit Laws-4
(2)limx→16
√x
x + 16=
√16
16+ 16=
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Exercises for Limit Laws-4
(2)limx→16
√x
x + 16=
√16
16+ 16= 432
=
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Exercises for Limit Laws-4
(2)limx→16
√x
x + 16=
√16
16+ 16= 432
= 18
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Exercises for Limit Laws-5
(3)limx→16
√x − 4x − 16
=
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Exercises for Limit Laws-5
(3)limx→16
√x − 4x − 16
=
limx→16
(√x − 4x − 16
)(√x + 4√x + 4
)=
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Exercises for Limit Laws-5
(3)limx→16
√x − 4x − 16
=
limx→16
(√x − 4x − 16
)(√x + 4√x + 4
)= limx→16
(√x)2 − 42
(x − 16)(√x + 4)
=
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Exercises for Limit Laws-5
(3)limx→16
√x − 4x − 16
=
limx→16
(√x − 4x − 16
)(√x + 4√x + 4
)= limx→16
(√x)2 − 42
(x − 16)(√x + 4)
=
limx→16
x − 16(x − 16)(
√x + 4)
=
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Exercises for Limit Laws-5
(3)limx→16
√x − 4x − 16
=
limx→16
(√x − 4x − 16
)(√x + 4√x + 4
)= limx→16
(√x)2 − 42
(x − 16)(√x + 4)
=
limx→16
x − 16(x − 16)(
√x + 4)
= limx→16
1√x + 4
=
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Exercises for Limit Laws-5
(3)limx→16
√x − 4x − 16
=
limx→16
(√x − 4x − 16
)(√x + 4√x + 4
)= limx→16
(√x)2 − 42
(x − 16)(√x + 4)
=
limx→16
x − 16(x − 16)(
√x + 4)
= limx→16
1√x + 4
= 1limx→16
(√x + 4)
=
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Exercises for Limit Laws-5
(3)limx→16
√x − 4x − 16
=
limx→16
(√x − 4x − 16
)(√x + 4√x + 4
)= limx→16
(√x)2 − 42
(x − 16)(√x + 4)
=
limx→16
x − 16(x − 16)(
√x + 4)
= limx→16
1√x + 4
= 1limx→16
(√x + 4)
= 1√16+ 4
=
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Exercises for Limit Laws-5
(3)limx→16
√x − 4x − 16
=
limx→16
(√x − 4x − 16
)(√x + 4√x + 4
)= limx→16
(√x)2 − 42
(x − 16)(√x + 4)
=
limx→16
x − 16(x − 16)(
√x + 4)
= limx→16
1√x + 4
= 1limx→16
(√x + 4)
= 1√16+ 4
= 14+ 4
=
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Exercises for Limit Laws-5
(3)limx→16
√x − 4x − 16
=
limx→16
(√x − 4x − 16
)(√x + 4√x + 4
)= limx→16
(√x)2 − 42
(x − 16)(√x + 4)
=
limx→16
x − 16(x − 16)(
√x + 4)
= limx→16
1√x + 4
= 1limx→16
(√x + 4)
= 1√16+ 4
= 14+ 4
= 18
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Exercises for Limit Laws-6
(4) limx→3
x − 3x2 − 9
=
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Exercises for Limit Laws-6
(4) limx→3
x − 3x2 − 9
= limx→3
x − 3(x − 3)(x + 3)
=
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Exercises for Limit Laws-6
(4) limx→3
x − 3x2 − 9
= limx→3
x − 3(x − 3)(x + 3)
= limx→3
1x + 3
=
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Exercises for Limit Laws-6
(4) limx→3
x − 3x2 − 9
= limx→3
x − 3(x − 3)(x + 3)
= limx→3
1x + 3
= 13+ 3
=
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Exercises for Limit Laws-6
(4) limx→3
x − 3x2 − 9
= limx→3
x − 3(x − 3)(x + 3)
= limx→3
1x + 3
= 13+ 3
= 16
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Exercises for Limit Laws-7
(5)limx→2
x2 − 4x − 2
=
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Exercises for Limit Laws-7
(5)limx→2
x2 − 4x − 2
= limx→2
(x + 2)(x − 2)x − 2
=
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Exercises for Limit Laws-7
(5)limx→2
x2 − 4x − 2
= limx→2
(x + 2)(x − 2)x − 2
= limx→2
(x + 2) =
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Exercises for Limit Laws-7
(5)limx→2
x2 − 4x − 2
= limx→2
(x + 2)(x − 2)x − 2
= limx→2
(x + 2) = 2+ 2 =
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Exercises for Limit Laws-7
(5)limx→2
x2 − 4x − 2
= limx→2
(x + 2)(x − 2)x − 2
= limx→2
(x + 2) = 2+ 2 = 4
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Exercises for Limit Laws-8
(6)limx→1−
⌊x − 12
⌋
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Exercises for Limit Laws-8
(6)limx→1−
⌊x − 12
⌋
Since x approaches 1 from the left,
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Exercises for Limit Laws-8
(6)limx→1−
⌊x − 12
⌋
Since x approaches 1 from the left, x − 1 will be negative,
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Exercises for Limit Laws-8
(6)limx→1−
⌊x − 12
⌋
Since x approaches 1 from the left, x − 1 will be negative, so⌊x − 12
⌋= −1 when x is in (−1,1).
Back to Questions
Exercises for Limit Laws-8
(6)limx→1−
⌊x − 12
⌋
Since x approaches 1 from the left, x − 1 will be negative, so⌊x − 12
⌋= −1 when x is in (−1,1). Thus
Back to Questions
Exercises for Limit Laws-8
(6)limx→1−
⌊x − 12
⌋
Since x approaches 1 from the left, x − 1 will be negative, so⌊x − 12
⌋= −1 when x is in (−1,1). Thus
limx→1−
⌊x − 12
⌋=
Back to Questions
Exercises for Limit Laws-8
(6)limx→1−
⌊x − 12
⌋
Since x approaches 1 from the left, x − 1 will be negative, so⌊x − 12
⌋= −1 when x is in (−1,1). Thus
limx→1−
⌊x − 12
⌋= limx→1−
(−1) =
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Exercises for Limit Laws-8
(6)limx→1−
⌊x − 12
⌋
Since x approaches 1 from the left, x − 1 will be negative, so⌊x − 12
⌋= −1 when x is in (−1,1). Thus
limx→1−
⌊x − 12
⌋= limx→1−
(−1) = −1
Back to Questions
Exercises for Limit Laws-9
(7) limx→1−
|x − 1|x − 1
=
Back to Questions
Exercises for Limit Laws-9
(7) limx→1−
|x − 1|x − 1
= limx→1−
1− xx − 1
=
Back to Questions
Exercises for Limit Laws-9
(7) limx→1−
|x − 1|x − 1
= limx→1−
1− xx − 1
= limx→1−
−1 =
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Exercises for Limit Laws-9
(7) limx→1−
|x − 1|x − 1
= limx→1−
1− xx − 1
= limx→1−
−1 = −1
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Exercises for Limit Laws-10
(8) limx→1+
|x − 1|x − 1
=
Back to Questions
Exercises for Limit Laws-10
(8) limx→1+
|x − 1|x − 1
= limx→1+
x − 1x − 1
=
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Exercises for Limit Laws-10
(8) limx→1+
|x − 1|x − 1
= limx→1+
x − 1x − 1
= limx→1+
1 =
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Exercises for Limit Laws-10
(8) limx→1+
|x − 1|x − 1
= limx→1+
x − 1x − 1
= limx→1+
1 = 1
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Exercises for Limit Laws-11
(9) limx→∞
x2 + 1+ 11− x =
Back to Questions
Exercises for Limit Laws-11
(9) limx→∞
x2 + 1+ 11− x = −∞
Back to Questions
Exercises for Limit Laws-12
(10)limx→−∞
x − 1x2 − 3x − 1
=
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Exercises for Limit Laws-12
(10)limx→−∞
x − 1x2 − 3x − 1
= 0
