LIMITS AT INFINITY

When we say limits at infinity, we mean one of the following limits:

𝐥𝐢𝐦 𝒇(𝒙) = 𝑳 𝐥𝐢𝐦 𝒇(𝒙) = 𝑳

𝒙→+∞ 𝒙→−∞

We are simply going to observe at what happens to a function if we let 𝑥 grows larger and
larger in either positive or negative sense. Remember that, the symbols +∞ and −∞ refer to
positive and negative infinity, respectively. These symbols do not represent real numbers but
they represent a very huge in magnitude of number at unbounded conditions.

LIMITS AT INFINITY THEOREMS

1. Let 𝑛 be any positive integer.

𝟏
a. 𝐥𝐢𝐦 =𝟎
𝒙→+∞ 𝒙𝒏

1
EXAMPLE: Evaluate lim
𝑥→+∞ 𝑥 2

1
To realize the theorem 1.a., observe the behavior of 𝑓(𝑥) = 𝑥 2 in the table as 𝑥 grows
positively infinite.

As the value of 𝒙 grows larger and larger, positively infinite,

𝑥 1 10 100 1000 10, 000

1
𝑓(𝑥) = 2 1 0.01 0.0001 0.000001 1𝑥10−8
𝑥

The value of 𝒇(𝒙) goes closer and closer to zero, 𝟎.

𝟏
Therefore, we can say that indeed, 𝐥𝐢𝐦 =𝟎
𝒙→+∞ 𝒙𝒏

𝟏
b. 𝐥𝐢𝐦 =𝟎
𝒙→−∞ 𝒏
𝒙

1
EXAMPLE: Evaluate lim
𝑥→−∞ 3
𝑥
 1
To realize the theorem 1.b., observe the behavior of 𝑓(𝑥) = 𝑥 3 in the table as 𝑥 grows
negatively infinite.

As the value of 𝒙 grows larger and larger in magnitude but has negative
sign, negatively infinite,

𝑥 −1 −10 −100 −1000 −10, 000

1
𝑓(𝑥) = 3 −1 −0.001 −0.000001 −0.000000001 −1𝑥10−12
𝑥

The value of 𝒇(𝒙) goes closer and closer to zero, 𝟎.

𝟏
Therefore, we can say that indeed, 𝐥𝐢𝐦 =𝟎
𝒙→−∞ 𝒙𝒏

Let’s solve problems using the theorem 1.a. and 1.b. Evaluate the following:

−3
1.1. lim
𝑥→∞ 3
2𝑥

SOLUTION:

−3 3 1
lim 3
= − lim 3
𝑥→∞ 2𝑥 2 𝑥→∞ 𝑥

3
=− (0)
2

=0

5
1.2. lim
𝑥→−∞ 𝑥 6

SOLUTION:

5 1
lim = 5 lim
𝑥→−∞ 𝑥 6 𝑥→−∞ 𝑥 6

= 5 (0)

=0
 4
1.3. lim
𝑥→−∞ 5𝑥 7

SOLUTION:

4 4 1
lim 7
= lim 7
𝑥→−∞ 5𝑥 5 𝑥→−∞ 𝑥

4
= (0)
5

=0

Now, after understanding the solution for 1.1., 1.2., and 1.3. Can you realize the theorem “Let 𝑛 be a
𝒌 𝒌
positive real number and 𝑘 is any real number except zero, then, 𝐥𝐢𝐦 = 𝟎 𝑎𝑛𝑑 𝐥𝐢𝐦 = 𝟎 "? This
𝒙→+∞ 𝒏
𝒙 𝒙→−∞ 𝒏
𝒙
theorem supports your answer in 1.1., 1.2., and 1.3. Thus,

𝒌 𝒌
𝐥𝐢𝐦 =𝟎 𝐥𝐢𝐦 =𝟎
𝒙→+∞ 𝒙𝒏 𝒙→−∞ 𝒙𝒏

2. Let 𝑛 be a positive real number and 𝑘 is any real number except zero.

a. lim 𝑥 𝑛 = +∞
𝑥→+∞

EXAMPLE A.1.: Evaluate 𝐥𝐢𝐦 𝒙𝟑

𝒙→+∞

To realize the theorem 2.a., observe the behavior of 𝑓(𝑥) = 𝑥 3 in the table as 𝑥 grows
positively infinite.

As the value of 𝒙 grows larger and larger, positively infinite,

𝑥 1 10 100 1000 10, 000

𝑓(𝑥) = 𝑥 3 1 1000 1,000,000 1,000,000,000 1𝑥1012

The value of 𝒇(𝒙) grows infinitely positive.

Therefore, we can say that indeed, 𝐥𝐢𝐦 𝒙𝒏 = +∞

𝒙→∞

EXAMPLE A.2.: Evaluate 𝐥𝐢𝐦 −𝟐𝒙𝟒

𝒙→+∞
 SOLUTION:

lim −2𝑥 4 = −2 lim 𝑥 4

𝑥→+∞ 𝑥→+∞

= −2(+∞)

= −∞

EXAMPLE A.3.: Evaluate 𝐥𝐢𝐦 𝟓𝒙𝟑

𝒙→+∞

SOLUTION:

lim 5𝑥 3 = 5 lim 𝑥 3
𝑥→+∞ 𝑥→+∞

= 5(+∞)

= +∞

+∞ 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
b. lim 𝑥 𝑛 = {
𝑥→−∞ −∞ 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑

EXAMPLE B.1.: Evaluate 𝐥𝐢𝐦 𝒙𝟑

𝒙→−∞

To realize the theorem 2.b., observe the behavior of 𝑓(𝑥) = 𝑥 3 in the table as 𝑥 grows
negatively infinite.

As the value of 𝒙 grows larger and larger in magnitude but has negative
sign, negatively infinite,

𝑥 −1 −10 −100 −1000 −10, 000

𝑓(𝑥) = 𝑥 3 −1 −1,000 −1,000,000 −1,000,000 −1𝑥1012

The value of 𝒇(𝒙) grows infinitely negative.

Therefore, we can say that indeed, 𝐥𝐢𝐦 𝒙𝒏 = +∞

𝒙→∞

EXAMPLE B.2.: Evaluate 𝐥𝐢𝐦 𝒙𝟒

𝒙→−∞
 To realize the theorem 2.d., observe the behavior of 𝑓(𝑥) = 𝑥 4 in the table as 𝑥 grows
negatively infinite.

As the value of 𝒙 grows larger and larger in magnitude but has negative
sign, negatively infinite,

𝑥 −1 −10 −100 −1000 −10, 000

𝑓(𝑥) = 𝑥 4 1 10,000 100,000,000 1𝑥1012 1𝑥1016

The value of 𝒇(𝒙) grows infinitely positive.

𝐥𝐢𝐦 𝒙𝒏 = +∞
Therefore, we can say that indeed, 𝒙→−∞

EXAMPLE B.3.: Evaluate 𝐥𝐢𝐦 −𝟐𝒙𝟓

𝒙→−∞
SOLUTION:

lim −2𝑥 5 = −2 lim 𝑥 5

𝑥→−∞ 𝑥→−∞

= −2(−∞)

= +∞

𝟑
EXAMPLE B.4.: Evaluate 𝐥𝐢𝐦 − 𝒙𝟔
𝒙→−∞ 𝟐

SOLUTION:

3 3
lim − 𝑥 6 = − lim 𝑥 6
𝑥→−∞ 2 2 𝑥→−∞

3
= − (+∞)
2

= −∞
 𝑓(𝑥)
3. Let ℎ(𝑥) = 𝑔(𝑥), where 𝑓(𝑥) and 𝑔(𝑥) are polynomials.

a. If the degree of 𝑓(𝑥) is less than the degree of 𝑔(𝑥), then,

𝒇(𝒙) 𝒇(𝒙)
𝐥𝐢𝐦 =𝟎 𝐥𝐢𝐦 =𝟎
𝒙→+∞ 𝒈(𝒙) 𝒙→−∞ 𝒈(𝒙)

b. If the degree of 𝑓(𝑥) is equal to the degree of 𝑔(𝑥), where, 𝑎 is the leading coefficient of
𝑓(𝑥) and 𝑏 is the leading coefficient of 𝑔(𝑥), then

𝒇(𝒙) 𝒂
𝐥𝐢𝐦 = 𝒇(𝒙) 𝒂
𝒙→+∞ 𝒈(𝒙) 𝒃 𝐥𝐢𝐦 =
𝒙→−∞ 𝒈(𝒙) 𝒃

c. If the degree of 𝑓(𝑥) is less than the degree of 𝑔(𝑥), then,

𝒇(𝒙) 𝒇(𝒙)
𝐥𝐢𝐦 = +∞ 𝒐𝒓 − ∞ 𝐥𝐢𝐦 = +∞ 𝒐𝒓 − ∞
𝒙→+∞ 𝒈(𝒙) 𝒙→−∞ 𝒈(𝒙)

𝑓(𝑥) 𝑓(𝑥)
In evaluating lim or lim , where 𝑓(𝑥) and 𝑔(𝑥) are polynomials, factor out the
𝑥→∞ 𝑔(𝑥) 𝑥→−∞ 𝑔(𝑥)
leading power of 𝑥.

𝟐𝒙𝟑 −𝒙+𝟑
EXAMPLE 3.A.: Evaluate 𝐥𝐢𝐦
𝒙→+∞ 𝒙𝟒 −𝟐𝒙𝟑 +𝒙𝟐 −𝒙+𝟕

SOLUTION:

1 3
2𝑥 3 − 𝑥 + 3 𝑥 3 (2 − 2 + 3 ) Factor out leading
lim = lim 𝑥 𝑥
𝑥→+∞ 𝑥 4 − 2𝑥 3 + 𝑥 2 − 𝑥 + 7 𝑥→+∞ 4 2 1 1 7 power of 𝑥.
𝑥 (1 − 𝑥 + 2 − 3 + 4 )
𝑥 𝑥 𝑥

1 3
(2 −
2 + 𝑥3) Remember
= lim 𝑥
𝑥→+∞ 2 1 1 7
𝑥 (1 − 𝑥 + 2 − 3 + 4 ) 𝒌
𝑥 𝑥 𝑥 𝐥𝐢𝐦 =𝟎
𝒙→+∞ 𝒙𝒏
 2 Fundamental Limit Theorem:
= lim
𝑥→+∞ 𝑥
lim 𝑘𝑓(𝑥) = klim 𝑓(𝑥)
𝑥→𝑐 𝑥→𝑐
1
= 2 lim
𝑥→+∞ 𝑥

= 2 (0) 𝟏
𝐥𝐢𝐦 =𝟎
𝒙→+∞ 𝒙𝒏
=0

Note: The answer also supports theorem 3.a.

𝟑𝒙𝟓 +𝒙𝟑 −𝟐𝒙+𝟏

EXAMPLE 3.B.: Evaluate 𝐥𝐢𝐦
𝒙→−∞ 𝟐𝒙𝟓 +𝟑𝒙𝟒 −𝟐𝒙𝟑 +𝟑𝒙𝟐 −𝟐𝒙+𝟓

SOLUTION:

1 2 1 Factor out
3𝑥 5 + 𝑥 3 − 2𝑥 + 1 𝑥 5 (3 + 2 − 4 + 5 ) leading power of
lim = lim 𝑥 𝑥 𝑥
𝑥→−∞ 2𝑥 5 + 3𝑥 4 − 2𝑥 3 + 3𝑥 2 − 2𝑥 + 5 𝑥→−∞ 5 3 2 3 2 5 𝑥.
𝑥 (2 + − 2 + 3 − 4 + 5 )
𝑥 𝑥 𝑥 𝑥 𝑥
Remember
1 2 1
(3 +
2 − 𝑥4 + 𝑥5)
𝒌 = lim 𝑥
𝐥𝐢𝐦 𝒏 = 𝟎 𝑥→−∞ 3 2 3 2 5
𝒙→+∞ 𝒙 (2 + 𝑥 − 2 + 3 − 4 + 5 )
𝑥 𝑥 𝑥 𝑥
Fundamental Limit Theorem 3
= lim
lim 𝑘 = 𝑘 𝑥→−∞ 2
𝑥→𝑐
3
=
2

Note: The answer also supports theorem 3.b.

−𝟑𝒙𝟖 +𝒙𝟑 −𝟐𝒙+𝟏

EXAMPLE 3.C.: Evaluate 𝐥𝐢𝐦
𝒙→−∞ 𝟐𝒙𝟓 +𝟑𝒙𝟒 −𝟐𝒙𝟑 +𝟑𝒙𝟐 −𝟐𝒙+𝟓

SOLUTION:

1 2 1
−3𝑥 8 + 𝑥 3 − 2𝑥 + 1 𝑥 8 (−3 + 2 − 4 + 5 ) Factor out
lim = lim 𝑥 𝑥 𝑥
𝑥→−∞ 2𝑥 5 + 3𝑥 4 − 2𝑥 3 + 3𝑥 2 − 2𝑥 + 5 𝑥→−∞ 5 3 2 3 2 5 leading power of
𝑥 (2 + − 2 + 3 − 4 + 5 ) 𝑥.
𝑥 𝑥 𝑥 𝑥 𝑥
Remember 1 2 1
𝑥 3 (−3 +
2 − 4 + 5)
𝒌 = lim 𝑥 𝑥 𝑥
𝐥𝐢𝐦 =𝟎 𝑥→−∞ 3 2 3 2 5
𝒙→+∞ 𝒙𝒏 (2 + − 2 + 3 − 4 + 5 )
𝑥 𝑥 𝑥 𝑥 𝑥
 Fundamental Limit Theorem: −3𝑥 3
= lim
𝑥→−∞ 2
lim 𝑘𝑓(𝑥) = klim 𝑓(𝑥)
𝑥→𝑐 𝑥→𝑐
3
=− lim 𝑥 3
2 𝑥→−∞

3
𝐥𝐢𝐦 𝒙𝒏 = −∞ 𝒊𝒇 𝒏 𝒊𝒔 𝒐𝒅𝒅 = − (−∞)
𝒙→−∞ 2

= +∞

Note: The answer also supports theorem 3.c.

It is now your turn to practice what you have learned. Solve the following practice problems.

MASTERY TEST.

Evaluate:

−2 3−𝑥 2 −𝑥 3
1. lim 7. lim 5𝑥 4 −3𝑥 3 +𝑥 2 −1
𝑥→∞ 3𝑥 5 𝑥→∞
11 5−2𝑥+4𝑥 2 −2𝑥 5
2. lim 8. lim
𝑥→−∞ −3𝑥 3 𝑥→−∞ 5𝑥 5 −4𝑥 3 +3𝑥 2 −𝑥+1
√2 3 −3𝑥 7
3. lim 9. lim 𝑥 5 −𝑥3 +𝑥2 −𝑥+1
𝑥→+∞ 𝑥 7 𝑥→∞
4. lim −2√3𝑥 4 √𝑥 2 +9
𝑥→−∞ 10. lim 𝑥+2
𝑥→−∞
𝑥3 8𝑦 6 −3𝑦
5. lim 15 11. lim
𝑥→−∞ 𝑦→−∞ 𝑦−11
2𝑥 2
6. lim 𝑥 6 −3𝑥−1
𝑥→+∞ 11 12. lim
𝑥→∞ 3−𝑥 6
 *****END OF DISCUSSION*****

If you have clarification, write in comment or message me at

obmontanijr@addu.edu.ph . Keep safe everyone.

STUDY. STUDY. STUDY

SOLVE. SOLVE. SOLVE.

PROCEED TO YOUR ASSIGNMENT

ANSWER TO MASTERY TEST

2
1. 0 2. 0 3. 0 4. −∞ 5. −∞ 6. +∞ 7. 0 8. − 9. −∞ 10. −1 11. −∞ 12. −1
5