Calculus 1 - Limits
Worksheet 10 – The
Squeezing Theorem
1
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Calculus 1 - Limits - Worksheet 10 – The Squeezing Theorem
1. Evaluate this limit using the Squeezing Theorem.
lim 𝑥 2 sin
𝑥→0
1
𝑥
2. Evaluate lim 𝑓(𝑥) using the Squeezing Theorem given that
𝑥→1
5 ≤ 𝑓(𝑥) ≤ 𝑥 2 + 6𝑥 − 2
2
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3. Evaluate this limit using the Squeezing Theorem.
lim 𝑥 4 sin
𝑥→0
7
𝑥
4. Evaluate this limit using the Squeezing Theorem.
lim 𝑥 2 cos
𝑥→0
3
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5
𝑥
5. Evaluate this limit using the Squeezing Theorem.
lim 𝑥 2 cos(10𝑥)
𝑥→0
6. Evaluate this limit using the Squeezing Theorem.
lim 𝑥 2 sin
𝑥→0
4
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5
+1
𝑥
7. Evaluate this limit using the Squeezing Theorem.
3
lim 𝑥 2 sin + 2
𝑥→0
𝑥
8. Evaluate this limit using the Squeezing Theorem.
lim 𝑥 2 sin
𝑥→0
5
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9
−2
𝑥
9. Evaluate this limit using the Squeezing Theorem.
lim 𝑥sin𝑥
𝑥→0
10. Evaluate this limit using the Squeezing Theorem.
lim 𝑥 2 sin𝑥
𝑥→0
6
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11. Evaluate this limit using the Squeezing Theorem.
lim 𝑥 2 sin
𝑥→0
1
𝑥2
12. Evaluate this limit using the Squeezing Theorem.
lim 𝑥 2 cos
𝑥→0
7
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1
𝑥2
Let’s switch gears and do three limits without the Squeezing Theorem. You might
have to use some imagination to do these problems.
13. Evaluate this limit.
1 − cos𝑥
𝑥→0
𝑥
lim
14. Evaluate this limit.
1 − cos𝑥
𝑥→0
sin𝑥
lim
15. Evaluate this limit.
sin(3𝑥)
𝑥→0 sin(4𝑥)
lim
8
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Answers - Calculus 1 - Limits - Worksheet 10 – The Squeezing Theorem
1. Evaluate this limit using the Squeezing Theorem.
lim 𝑥 2 sin
𝑥→0
1
𝑥
1
We know that −1 ≤ sin ≤ 1. Next, we can multiply this inequality by 𝑥 2
𝑥
without changing its correctness. Now we have
1
−𝑥 2 ≤ 𝑥 2 sin ≤ 𝑥 2
𝑥
Take the limit of each part of the inequality.
1
lim( − 𝑥 2 ) ≤ lim 𝑥 2 sin ≤ lim 𝑥 2
𝑥→0
𝑥→0
𝑥 𝑥→0
Next, we know that lim𝑥→0 ( − 𝑥 2 ) = 0 and lim𝑥→0 ( 𝑥 2 ) = 0. Thus, we have
1
0 ≤ lim 𝑥 2 sin ≤ 0
𝑥→0
𝑥
1
So, we can conclude that lim 𝑥 2 sin = 0
𝑥→0
𝑥
Graphically, the function squeezed between 𝑦 = −𝑥 2 and 𝑦 = 𝑥 2 shows the
same limit.
Answer: lim𝑥→0 𝑥 2 sin
1
𝑥
=0
9
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2. Evaluate lim 𝑓(𝑥) using the Squeezing Theorem given that
𝑥→1
5 ≤ 𝑓(𝑥) ≤ 𝑥 2 + 6x − 2
Since we are given the inequality, take the limits.
lim5 ≤ lim 𝑓(𝑥) ≤ lim( 𝑥 2 + 6𝑥 − 2)
𝑥→1
𝑥→1
𝑥→1
5 ≤ lim 𝑓(𝑥) ≤ ( 12 + 6(1) − 2)
𝑥→1
5 ≤ lim 𝑓(𝑥) ≤ 5
𝑥→1
Therefore, we can conclude that lim𝑓(𝑥) = 5
𝑥→1
Answer: lim𝑓(𝑥) = 5
𝑥→1
10
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3. Evaluate this limit using the Squeezing Theorem.
lim 𝑥 4 sin
𝑥→0
7
𝑥
7
We know that −1 ≤ sin ≤ 1. Next, we can multiply this inequality by 𝑥 4
𝑥
without changing its correctness. Now we have
7
−𝑥 4 ≤ 𝑥 4 sin ≤ 𝑥 4
𝑥
Take the limit of each part of the inequality.
7
lim( − 𝑥 4 ) ≤ lim 𝑥 4 sin ≤ lim 𝑥 4
𝑥→0
𝑥→0
𝑥 𝑥→0
Next, we know that lim𝑥→0 ( − 𝑥 4 ) = 0 and lim𝑥→0 ( 𝑥 4 ) = 0. Thus, we have
7
0 ≤ lim 𝑥 4 sin ≤ 0
𝑥→0
𝑥
7
So, we can conclude that lim 𝑥 4 sin = 0
𝑥→0
𝑥
Graphically, the function squeezed between 𝑦 = −𝑥 4 and 𝑦 = 𝑥 4 shows the
same limit.
7
Answer: lim𝑥 4 sin = 0
𝑥→0
𝑥
11
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4. Evaluate this limit using the Squeezing Theorem.
lim 𝑥 2 cos
𝑥→0
5
𝑥
1
We know that −1 ≤ cos ≤ 1. Next, we can multiply this inequality by 𝑥 2
𝑥
without changing its correctness. Now we have
5
−𝑥 2 ≤ 𝑥 2 cos ≤ 𝑥 2
𝑥
Take the limit of each part of the inequality.
5
lim ( − 𝑥 2 ) ≤ lim 𝑥 2 cos ≤ lim𝑥 2
𝑥→0
𝑥→0
𝑥 𝑥→0
Next, we know that lim𝑥→0 ( − 𝑥 2 ) = 0 and lim𝑥→0 ( 𝑥 2 ) = 0. Thus, we have
5
0 ≤ lim 𝑥 2 cos ≤ 0
𝑥→0
𝑥
5
So, we can conclude that lim 𝑥 2 cos = 0
𝑥→0
𝑥
Graphically, the function squeezed between 𝑦 = −𝑥 2 and 𝑦 = 𝑥 2 shows the
same limit.
5
Answer: lim𝑥→0 𝑥 2 cos = 0
𝑥
12
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5. Evaluate this limit using the Squeezing Theorem.
lim 𝑥 2 cos(10𝑥)
𝑥→0
We know that −1 ≤ cos(10𝑥) ≤ 1. Next, we can multiply this inequality by 𝑥 2
without changing its correctness. Now we have
−𝑥 2 ≤ 𝑥 2 cos(10𝑥) ≤ 𝑥 2
Take the limit of each part of the inequality.
lim( − 𝑥 2 ) ≤ lim𝑥 2 cos(10𝑥) ≤ lim 𝑥 2
𝑥→0
𝑥→0
𝑥→0
Next, we know that lim𝑥→0 ( − 𝑥 2 ) = 0 and lim𝑥→0 ( 𝑥 2 ) = 0. Thus, we have
0 ≤ lim𝑥 2 cos(10𝑥) ≤ 0
𝑥→0
So, we can conclude that lim 𝑥 2 cos(10𝑥) = 0
𝑥→0
Graphically, the function squeezed between 𝑦 = −𝑥 2 and 𝑦 = 𝑥 2 shows the
same limit.
Answer: lim𝑥 2 cos(10𝑥) = 0
𝑥→0
13
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6. Evaluate this limit using the Squeezing Theorem.
lim 𝑥 2 sin
𝑥→0
5
+1
𝑥
5
We know that −1 ≤ sin ≤ 1, so we also know that Next, we can multiply the
𝑥
2
inequality by 𝑥 and still preserve the inequality. Now we have
5
−𝑥 2 ≤ 𝑥 2 sin ≤ 𝑥 2 Next, we can add 1 to all parts of the inequality according to
𝑥
the addition property of inequalities. This gives us
5
−𝑥 2 + 1 ≤ 𝑥 2 sin + 1 ≤ 𝑥 2 + 1
𝑥
The inequality matches this problem. Take the limit of each part of the inequality.
5
lim(−𝑥 2 + 1) ≤ lim (𝑥 2 sin + 1) ≤ lim(𝑥 2 + 1)
𝑥→0
𝑥→0
𝑥→0
𝑥
Next, we know that lim𝑥→0 ( − 𝑥 2 + 1) = 1 and lim𝑥→0 ( 𝑥 2 + 1) = 1. Thus, we
have
5
1 ≤ lim (𝑥 2 sin + 1) ≤ 1
𝑥→0
𝑥
5
So, we can conclude that lim (𝑥 2 sin + 1) = 1
𝑥→0
𝑥
Graphically, the function squeezed between 𝑦 = −𝑥 2 + 1 and 𝑦 = 𝑥 2 + 1 shows
the same limit.
5
Answer: lim (𝑥 2 sin + 1) = 1
𝑥→0
𝑥
14
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7. Evaluate this limit using the Squeezing Theorem.
3
lim 𝑥 2 sin + 2
𝑥→0
𝑥
3
We know that −1 ≤ sin ≤ 1, so we also know that Next, we can multiply the
𝑥
inequality by 𝑥 2 and still preserve the inequality. Now we have
3
−𝑥 2 ≤ 𝑥 2 sin ≤ 𝑥 2 Next, we can add 2 to all parts of the inequality according to
𝑥
the addition property of inequalities. This gives us
5
−𝑥 2 + 2 ≤ 𝑥 2 sin + 2 ≤ 𝑥 2 + 2
𝑥
The inequality matches this problem. Take the limit of each part of the inequality.
3
lim(−𝑥 2 + 2) ≤ lim (𝑥 2 sin + 2) ≤ lim(𝑥 2 + 2)
𝑥→0
𝑥→0
𝑥→0
𝑥
Next, we know that lim𝑥→0 ( − 𝑥 2 + 2) = 2 and lim𝑥→0 ( 𝑥 2 + 2) = 2. Thus, we
have
3
2 ≤ lim (𝑥 2 sin + 2) ≤ 2
𝑥→0
𝑥
3
So, we can conclude that lim (𝑥 2 sin + 2) = 2
𝑥→0
𝑥
Graphically, the function squeezed between 𝑦 = −𝑥 2 + 2 and 𝑦 = 𝑥 2 + 2 shows
the same limit.
3
Answer: lim (𝑥 2 sin + 2) = 2
𝑥→0
𝑥
15
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8. Evaluate this limit using the Squeezing Theorem.
lim 𝑥 2 sin
𝑥→0
9
−2
𝑥
9
We know that −1 ≤ sin ≤ 1, so we also know that Next, we can multiply the
𝑥
inequality by 𝑥 2 and still preserve the inequality. Now we have
9
−𝑥 2 ≤ 𝑥 2 sin ≤ 𝑥 2 Next, we can subtract 2 from all parts of the inequality
𝑥
according to the subtraction property of inequalities. This gives us
9
−𝑥 2 − 2 ≤ 𝑥 2 sin − 2 ≤ 𝑥 2 − 2
𝑥
The inequality matches this problem. Take the limit of each part of the inequality.
9
lim(−𝑥 2 − 2) ≤ lim (𝑥 2 sin − 2) ≤ lim(𝑥 2 − 2)
𝑥→0
𝑥→0
𝑥→0
𝑥
Next, we know that lim𝑥→0 ( − 𝑥 2 − 2) = −2 and lim𝑥→0 ( 𝑥 2 − 2) = −2. Thus,
we have
9
−2 ≤ lim (𝑥 2 sin − 2) ≤ −2
𝑥→0
𝑥
9
So, we can conclude that lim (𝑥 2 sin − 2) = −2
𝑥→0
𝑥
Graphically, the function squeezed between 𝑦 = −𝑥 2 − 2 and 𝑦 = 𝑥 2 − 2 shows
the same limit.
9
Answer: lim (𝑥 2 sin − 2) = −2
𝑥→0
𝑥
16
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9. Evaluate this limit using the Squeezing Theorem.
lim 𝑥sin𝑥
𝑥→0
We know that −1 ≤ sin𝑥 ≤ 1, so we also know that Next, we can multiply the
inequality by the absolute value of 𝑥 and still preserve the inequality. Now we
have
−|𝑥| ≤ |𝑥|sin𝑥 ≤ |𝑥|
The inequality matches this problem, but the absolute value symbol is not
necessary in the center of the inequality. Take the limit of each part of the
inequality.
lim(−|𝑥|) ≤ lim(𝑥sin𝑥) ≤ lim(|𝑥|)
𝑥→0
𝑥→0
𝑥→0
Next, we know that lim𝑥→0 (− |𝑥|) = 0 and lim𝑥→0 ( |𝑥|) = 0. Thus, we have
0 ≤ lim (𝑥sin𝑥) ≤ 0
𝑥→0
So, we can conclude that lim (𝑥sin𝑥) = 0
𝑥→0
Graphically, the function squeezed between 𝑦 = −|𝑥| and 𝑦 = |𝑥| shows the
same limit.
Answer: lim(𝑥sin𝑥) = 0
𝑥→0
17
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10. Evaluate this limit using the Squeezing Theorem.
lim 𝑥 2 sin(12𝑥)
𝑥→0
We know that −1 ≤ sin(12𝑥) ≤ 1, so we also know that Next, we can multiply
the inequality by 𝑥 2 and still preserve the inequality. Now we have
−𝑥 2 ≤ 𝑥 2 sin(12𝑥) ≤ 𝑥 2
Take the limit of each part of the inequality.
lim(−𝑥 2 ) ≤ lim(𝑥 2 sin(12𝑥)) ≤ lim (𝑥 2 )
𝑥→0
𝑥→0
𝑥→0
Next, we know that lim𝑥→0 ( − 𝑥 2 ) = 0 and lim𝑥→0 ( 𝑥 2 ) = 0. Thus, we have
0 ≤ lim (𝑥 2 sin(12𝑥)) ≤ 0
𝑥→0
So, we can conclude that lim (𝑥 2 sin(12𝑥)) = 0
𝑥→0
Graphically, the function squeezed between 𝑦 = −𝑥 2 and 𝑦 = 𝑥 2 shows the
same limit.
Answer: lim(𝑥 2 sin(12𝑥)) = 0
𝑥→0
18
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11. Evaluate this limit using the Squeezing Theorem.
lim 𝑥 2 sin
𝑥→0
We know that −1 ≤ sin
1
𝑥2
1
𝑥2
≤ 1, so we also know that Next, we can multiply the
inequality by 𝑥 2 and still preserve the inequality. Now we have
−𝑥 2 ≤ 𝑥 2 sin
1
≤ 𝑥2
𝑥2
Take the limit of each part of the inequality.
lim (−𝑥 2 ) ≤ lim (𝑥 2 sin
𝑥→0
𝑥→0
1
) ≤ lim (𝑥 2 )
2
𝑥→0
𝑥
Next, we know that lim𝑥→0 ( − 𝑥 2 ) = 0 and lim𝑥→0 ( 𝑥 2 ) = 0. Thus, we have
0 ≤ lim (𝑥 2 sin
𝑥→0
1
)≤0
𝑥2
1
So, we can conclude that lim (𝑥 2 sin 2) = 0
𝑥
𝑥→0
Graphically, the function squeezed between 𝑦 = −𝑥 2 and 𝑦 = 𝑥 2 shows the
same limit.
1
Answer: lim (𝑥 2 sin 2) = 0
𝑥
𝑥→0
19
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12. Evaluate this limit using the Squeezing Theorem.
lim 𝑥 2 cos
𝑥→0
We know that −1 ≤ cos
1
𝑥2
1
𝑥2
≤ 1, so we also know that Next, we can multiply the
inequality by 𝑥 2 and still preserve the inequality. Now we have
−𝑥 2 ≤ 𝑥 2 cos
1
≤ 𝑥2
𝑥2
Take the limit of each part of the inequality.
lim (−𝑥 2 ) ≤ lim (𝑥 2 cos
𝑥→0
𝑥→0
1
) ≤ lim(𝑥 2 )
2
𝑥→0
𝑥
Next, we know that lim𝑥→0 ( − 𝑥 2 ) = 0 and lim𝑥→0 ( 𝑥 2 ) = 0. Thus, we have
0 ≤ lim (𝑥 2 cos
𝑥→0
1
)≤0
𝑥2
1
So, we can conclude that lim (𝑥 2 cos 2) = 0
𝑥
𝑥→0
Graphically, the function squeezed between 𝑦 = −𝑥 2 and 𝑦 = 𝑥 2 shows the
same limit.
1
Answer: lim (𝑥 2 cos 2) = 0
𝑥
𝑥→0
20
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13. Evaluate this limit.
1 − cos𝑥
𝑥→0
𝑥
lim
If we try to evaluate the limit using direct substitution we get an indeterminate
answer. Conjugate the numerator.
1 − cos𝑥
1 − cos𝑥 1 + cos𝑥
1 − cos 2 𝑥
lim
= lim
∙
= lim
𝑥→0
𝑥→0
𝑥
𝑥
1 + cos𝑥 𝑥→0 𝑥 + 𝑥cos𝑥
Use the trig identity: sin2 𝑥 + cos 2 𝑥 = 1 which means sin2 𝑥 = 1 − cos 2 𝑥.
1 − cos 2 𝑥
sin2 𝑥
lim
= lim
𝑥→0 𝑥 + 𝑥cos𝑥
𝑥→0 𝑥(1 + cos𝑥)
Split the limit into a product of two fractions and then into the product of two
limits.
sin2 𝑥
sin𝑥
𝑠𝑖𝑛𝑥
sin𝑥
𝑠𝑖𝑛𝑥
= lim
∙
= lim
∙ lim
𝑥→0 𝑥(1 + cos𝑥)
𝑥→0 𝑥
1 + cos𝑥 𝑥→0 𝑥 𝑥→0 1 + cos𝑥
lim
Remember from a video from the beginning of the Limit Series, lim𝑥→0
sin𝑥
𝑠𝑖𝑛𝑥
𝑠𝑖𝑛𝑥
∙ lim
= 1 ∙ lim
𝑥→0 𝑥
𝑥→0 1 + cos𝑥
𝑥→0 1 + cos𝑥
lim
Now, evaluate using direct substitution.
𝑠𝑖𝑛𝑥
𝑠𝑖𝑛0
0
=
= =0
𝑥→0 1 + cos𝑥
1 + cos0 2
lim
Answer: lim𝑥→0
1−cos𝑥
𝑥
=0
21
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sin𝑥
𝑥
= 1 so
14. Evaluate this limit.
1 − cos𝑥
𝑥→0
sin𝑥
lim
Split the limit into the limit of the product of two fractions.
1 − cos𝑥
1 1 − cos𝑥
= lim
∙
𝑥→0
𝑥→0 sin𝑥
sin𝑥
1
lim
𝑥
Manipulate the fractions by multiplying both of them by .
𝑥
1 1 − cos𝑥
𝑥
𝑥(1 − cos𝑥)
∙
= lim
∙
𝑥→0 sin𝑥
𝑥→0 𝑥sin𝑥
1
𝑥
lim
Factor an 𝑥 from the denominator of the first fraction and from the numerator of
the second fraction. Then, cancel those 𝑥′s.
𝑥 𝑥 (1 − cos𝑥)
𝑥 (1 − cos𝑥)
lim ∙
∙
= lim
∙
𝑥→0 𝑥 sin𝑥
𝑥→0 sin𝑥
𝑥
𝑥
We already know that lim𝑥→0
lim𝑥→0
1−cos𝑥
𝑥
sin𝑥
𝑥
= 1 and in the last problem we proved that
= 0. Therefore, lim𝑥→0
Answer: lim𝑥→0
1−cos𝑥
sin𝑥
𝑥
sin𝑥
=0
22
© MathTutorDVD.com
∙
(1−cos𝑥)
𝑥
= 1 ∙ 0 = 0.
15. Evaluate this limit.
sin(3𝑥)
𝑥→0 sin(4𝑥)
lim
We will use the fact that lim𝑥→0
sin𝑥
𝑥
= 1 so lim𝑥→0
sin𝑢
𝑢
= 1.
Separate the limit into the limit of the product of two fractions and then modify
the fractions to match the above limits.
sin(3𝑥)
sin(3𝑥)
1
3𝑥sin(3𝑥)
4𝑥
= lim
∙
= lim
∙
𝑥→0 sin(4𝑥)
𝑥→0
1
sin(4𝑥) 𝑥→0
3𝑥
4𝑥sin(4𝑥)
lim
Factor out the 3𝑥 in the numerator of the first fraction and the 4𝑥 in the
denominator of the second fraction.
3𝑥 sin(3𝑥)
4𝑥
∙
∙
𝑥→0 4𝑥
3𝑥
sin(4𝑥)
lim
Cancel
𝑥
𝑥
, factor out
3
4
and rewrite the limit of the product as a product of two
limits.
3𝑥 sin(3𝑥)
4𝑥
3
sin(3𝑥)
4𝑥
∙
∙
= lim
∙ lim
𝑥→0 4𝑥
𝑥→0 sin(4𝑥)
3𝑥
sin(4𝑥) 4 𝑥→0 3𝑥
lim
Evaluate the two limits based on our knowledge of those limits.
3
sin(3𝑥)
4𝑥
3
3
lim
∙ lim
= ∙1∙1=
𝑥→0 sin(4𝑥)
4 𝑥→0 3𝑥
4
4
Answer: lim𝑥→0
sin(3𝑥)
sin(4𝑥)
=
3
4
23
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