Limits — problems and solutions

brought to you by sciency.tech

last updated: February 13, 2019

Summary: This document contains some of the most common limits problems for you to
review! Feel free to jump around or start from the beginning! Visit https://sciency.tech for
the solutions and other problem-and-solution guides!

Contents
1 How to read limits out loud 2

2 Basic limit problems 3

3 One-sided limits 5

4 Limit laws 7

5 Harder limit problems 10

6 l’Hôpital’s rule 14

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How to read limits out loud sciency.tech

1 How to read limits out loud

1. How do you read f (x)?

Solution: “F” of “X.”

2. How do you read lim f (x) = L?

x→a

Solution: The limit of “F” as “X” approaches “A” is “L.”

3. How do you read lim− f (x)?

x→a

Solution: The limit of “F” as “X” approaches “A” from the left.

4. How do you read lim+ f (x)?

x→a

Solution: The limit of “F” as “X” approaches “A” from the right.

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Basic limit problems sciency.tech

2 Basic limit problems

1. lim x = ?
x→3

Solution:
lim x = 3.
x→3

2. lim (x2 + 7) = ?
x→a

Solution:
x2 + 7 = a2 + 7.


lim
x→a

x
3. lim cos =?
x→π 2

Solution:
x π 
lim cos = cos
x→π 2 2
= 1.

4. lim e−x = ?
x→∞

Solution:

lim e−x = e−∞

x→∞
= 1.

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Basic limit problems sciency.tech

x−3
5. lim =?
x→a x2 + 7

Solution:
x−3 a−3
lim = .
x→a x2 + 7 a2 + 7

6. lim x cos x = ?
x→π

Solution:

lim x cos x = π cos π

x→π
= −π.

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One-sided limits sciency.tech

3 One-sided limits
1. Let (
x + 2, if x < 0
f (x) = ,
3x − 7, if x ≥ 0
then
lim f (x) = ?
x→0+

Solution:

lim f (x) = lim (3x − 7)

x→0+ x→0+
= 0−7
= 7.

2. Let (
x + 2, if x < 0
f (x) =
3x − 7, if x ≥ 0
then
lim f (x) = ?
x→0−

Solution:

lim f (x) = lim (x + 2)

x→0− x→0−
= 0+2
= 2.

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One-sided limits sciency.tech

3. Let (
x + 2, if x < 0
f (x) = ,
3x − 7, if x ≥ 0
then
lim f (x) = ?
x→0

Solution:
lim f (x) does not exist
x→0

since
lim f (x) 6= lim− f (x).
x→0+ x→0

Recall from the previous two questions that

lim f (x) = 2
x→0−

and
lim f (x) = −7.
x→0+

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Limit laws sciency.tech

4 Limit laws
1. lim (f (x) + g(x)) = ?
x→a

Solution: lim (f (x) + g(x)) = lim f (x) + lim g(x).

x→a x→a x→a

2. lim f (x) g(x) = ?

x→a

Solution:   
lim (f (x) g(x)) = lim f (x) lim g(x) .
x→a x→a x→a

f (x)
3. lim =?
x→a g(x)

Solution:
f (x) lim f (x)
lim = x→a .
x→a g(x) lim g(x)
x→a

4. lim f (g(x)) = ?
x→a

Solution:  
lim f (g(x)) = f lim g(x) ,
x→a x→a

assuming f is a continuous function.

5. lim 17 = ?
x→a

Solution:
lim 17 = 17.
x→a

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Limit laws sciency.tech

6. lim (f (x))2 = ?
x→a

Solution:  2
2
lim (f (x)) = lim f (x) .
x→a x→a

7. lim (f (x))n = ?
x→a

Solution:  n
lim (f (x))n = lim f (x) .
x→a x→a

8. lim (7x − 2)3 = ?

x→a

Solution:
 3
3
lim (7x − 2) = lim (7x − 2)
x→a x→a
= (7a − 2)3 .

√
9. lim x+4= ?
x→0

Solution: Since the square root function is continuous,

√ q
lim x+4 = lim (x + 4)
x→0 x→0
√
= 4
= 2.

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Limit laws sciency.tech
√
10. lim x+4= ?
x→ −7

Solution:
√ √
lim x+4 = −7 + 4
x→ −7
√
= −3
√
= i 3,
√
where i = −1 is called the ”imaginary number.”

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Harder limit problems sciency.tech

5 Harder limit problems

x2 − 25
1. lim =?
x→5 x−5

Solution:
x2 − 25 (x + 5)(x − 5)
lim = lim
x→5 x−5 x→5 (x − 5)
= lim (x + 5)
x→5
= 5+5
= 10.

1
2. lim =?
x→∞ x

Solution:
1 1
lim =
x→∞ x lim x
x→∞
1
=
∞
= 0.

1
3. lim =?
x→∞ x2

Solution:
1 1
lim =
x→∞ x2 lim x2
x→∞
1
=
∞
= 0.

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Harder limit problems sciency.tech

4. Let an = 2 + 1/n. Then lim an = ?

n→∞

Solution:
 
1
lim an = lim 2 +
n→∞ n→∞ n
1
= lim 2 + lim
n→∞ n→∞ n
= 2+0
= 2.

√
x2 + 49 − 7
5. lim =?
x→0 x2

0
Solution: First, notice that directly plugging 0 into x gives us , so we need another
0
way.
If we instead multiply the top and bottom by
√
x2 + 49 + 7,

then we get
√ √
x2 + 49 − 7 x2 + 49 + 7
lim · √
x→0 x2 x2 + 49 + 7
(x2 + 49) − 49
= lim √

x→0 x2 x2 + 49 + 7
x2
= lim √

x→0 x2 x2 + 49 + 7
1
= lim √
x→0 x2 + 49 + 7
1
=
7+7
1
= .
14

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Harder limit problems sciency.tech
√
x2 + 24 − 7
6. lim =?
x→5 x2 − 25

0
Solution: Similar to the previous problem, directly plugging 5 into x gives us
,
0
so we need to use another approach. We try multiplying the top and bottom by
√
x2 + 24 + 7:
√ √
x2 + 24 − 7 x2 + 24 + 7
lim · √
x→5 x2 − 25 x2 + 24 + 7
(x2 + 24) − 49
= lim √
x→5 (x2 − 25)( x2 + 24 + 7)

(x2 − 25)
= lim √
x→5 (x2 − 25)( x2 + 24 + 7)

1
= lim √
x→5 2
x + 24 + 7
1
= √
25 + 24 + 7
1
= .
14

√
4x4 + 24x − 7
7. lim =?
x→∞ x2 − 25
(Try calculating this limit without using l’Hôpital’s rule.)

Solution: First, divide the numerator and denominator by x2 and rewrite the limit
in terms of h = 1/x. To do this, understand that letting x → ∞ is the same as
1
letting h = → 0+ (see Note 2 for more details). This allows us to rewrite the limit
x
as
√ √
4x4 + 24x − 7 1/x2 4 + 24h3 − 7h4
lim · = lim
x→∞ x2 − 25 1/x2 h→0+ 1 − 25h2
√
4
=
1
= 2.

Note 1:
Since both the numerator and denominator approaches infinity, you may be tempted

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Harder limit problems sciency.tech

to use l’Hôpital’s rule. Of course, you can and that is the standard approach, but it
will be messy. Give it a try and decide for yourself which approach is better.

Note 2:
The new limit is for h approaching zero from the right-hand side. This is because we
require h to remain positive (h > 0) at all times since x remained positive as x ap-
proached infinity. However, it actually doesn’t matter here since both the numerator
and denominator are continuous at h = 0.

5x3 + 4x + 7
8. lim =?
x→−∞ 25 − 2x3

(Try calculating this limit without using l’Hôpital’s rule.)

−1
Solution: First, divide both the numerator and denominator by x3 and let h = .
x
This allows us to rewrite the limit as
5x3 + 4x + 7 5 + 4h2 − 7h3
lim = lim+
x→−∞ 25 − 2x3 h→0 25h3 − 2
−5
= .
2

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l’Hôpital’s rule sciency.tech

6 l’Hôpital’s rule
1. What is l’Hôpital’s rule?

Solution: L’Hôpital’s rule is a method that lets us use derivatives in evaluating

limits involving ”indeterminate forms,” i.e. when a straight-forward approach gives
0 ±∞
us or .
0 ±∞

More specificially, l’Hôpital’s rule tells us that when

lim f (x) 0 ±∞
x→a
= or ,
lim g(x) 0 ±∞
x→a

f (x) f 0 (x)
lim = lim 0 ,
x→a g(x) x→a g (x)
where the primes ( ’ ) signify taking the derivative with respect to x.

sin x
2. lim =?
x→0 x

sin 0 0
Solution: Since = , we apply l’Hôpital’s rule:
0 0
sin x (sin x)0
lim = lim
x→0 x x→0 (x)0
cos x
= lim
x→0 1
= cos 0
= 1.

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l’Hôpital’s rule sciency.tech

(x + 3)3
3. lim =?
x→−3 x2 + 9

Solution: Since
lim (x + 3)3 0
x→−3
= ,
lim (x2 + 9) 0
x→−3

we apply l’Hôpital’s rule:

(x + 3)3 3(x + 3)2

lim = lim
x→−3 x2 + 9 x→−3 2x
0
=
−6
= 0.

ex
4. lim =?
x→∞ x2 + 4

Solution: Since
lim ex ∞
x→∞
= ,
lim (x2 + 4) ∞
x→∞
we apply l’Hôpital’s rule, which gives us
ex (ex )0
lim 2 = lim
x→∞ x + 4 x→∞ (x2 + 4)0
ex
= lim .
x→∞ 2x

Since
lim ex ∞
x→∞
= ,
lim 2x ∞
x→∞
we apply l’Hôpital’s rule a second time:
ex (ex )0
lim = lim
x→∞ 2x x→∞ (2x)0
ex
= lim
x→∞ 2
∞
=
2
= ∞.

And so, the answer is

ex
lim = ∞.
x→∞ x2 + 4

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