Limits at Infinity & Horizontal Asymptotes
Section 2.6
Prepared by:
Prof. Wafik Lotfallah
Prof. Maged G. Iskander
Dr. Kamal Tawfik
Lecture 8 Objectives
Evaluate Limits of functions at both ∞ and −∞.
Find Horizontal Asymptotes of function graphs.
Limits at infinity
Let’s begin by investigating the behavior of the function
𝑓 defined by
as 𝑥 becomes large.
As 𝑥 grows larger and larger you can see
that the values of 𝒇(𝒙) get closer and
closer to 𝟏 (as indicated in the table).
This situation is expressed symbolically by writing
In general, we use the notation
to indicate that the values of 𝑓(𝑥) approach 𝐿 as 𝑥 becomes
larger and larger.
Definition 𝟏
Geometric illustrations of Definition 1 are shown in Figure 2
Note that:
the line 𝑦 = 𝐿 is called a horizontal asymptote.
Limits at −∞
Referring back to Figure 1, we see that for numerically large
Negative values of 𝒙, the values of 𝑓(𝑥) are close to 1. This is
expressed by writing
Example 1
For the function 𝑓 whose graph is given, state the following.
Example 2
1 1
Find lim and lim .
𝑥→∞ 𝑥 𝑥→−∞ 𝑥
In fact, by taking 𝑥 large enough, we can
make 1Τ𝑥 close to 0 as we please.
Therefore, we have
1
lim =0.
𝑥→∞ 𝑥
Similar reasoning shows that when 𝑥 is
large negative, 1Τ𝑥 is small negative, so
we also have
1
lim =0.
𝑥→−∞ 𝑥
It follows that the line 𝑦 = 0 (the 𝑥-axis) is a horizontal
asymptote of the curve 𝑦 = 1Τ𝑥 .
In General:
Example 3
Evaluate
Hint:
In the case of evaluating a limit (Horizontal Asymptotes)
of a quotient of two polynomials, we divide by 𝒙 with
the highest power in the denominator.
Horizontal Asymptotes
Example: Find all horizontal asymptotes for the
function:
2x +1 2
f ( x) =
3x − 5
Dividing both numerator and denominator
by x and using the properties of limits,
we have:
1
2+ 2
2x + 1
2
x
lim = lim (since x 2 = x for x 0)
x → 3 x − 5 x → 5
3−
x
1 1
lim 2 + 2 lim 2 + lim 2
x → x x → x → x 2+0 2
= = = =
5 1 3 − 5.0 3
lim 3 − lim 3 − 5lim
x →
x x → x → x
In computing the limit as x → −
we must remember that, for x < 0,
we have
x = x = −x
2
So, when we divide the numerator by x, for x < 0,
we get
1 1 1
2x2 + 1 = − 2x2 + 1 = − 2 +
x x2 x2
Therefore,
1 1
2𝑥 2+1 − 2+ − 2 + lim 2
𝑥2 𝑥→−∞ 𝑥 2
lim = lim = =−
𝑥→−∞ 3𝑥 − 5 𝑥→−∞ 5 1 3
3− 3 − 5 lim
𝑥 𝑥→−∞ 𝑥
𝒙
Horizontal asymptote of 𝒆
The graph of the natural exponential function 𝑦 = 𝑒 𝑥 has the
line 𝑦 = 0 (the 𝑥-axis) as a horizontal asymptote.
Note that:
▪ lim 𝑒 𝑥 = lim 𝑒 −𝑥 = 0 . 𝐲 = 𝒆−𝒙
𝑥→−∞ 𝑥→∞
▪ lim 𝑒 𝑥 = lim 𝑒 −𝑥 = ∞ .
𝑥→∞ 𝑥→−∞
Example 4
Find the limit or show that it does not exist.
1 − 𝑒𝑥
1. lim 𝑥
.
𝑥→∞ 1 + 2𝑒
𝑒 3𝑥 − 𝑒 −3𝑥
2. lim 3𝑥 −3𝑥
.
𝑥→∞ 𝑒 + 𝑒
𝑒 3𝑥 − 𝑒 −3𝑥
3. lim 3𝑥 −3𝑥
.
𝑥→−∞ 𝑒 + 𝑒
4. lim [ ln 2 + 𝑥 − ln(1 + 𝑥) ] .
𝑥→∞
Solution
1 − 𝑒𝑥
1. lim .
𝑥→∞ 1 + 2𝑒 𝑥
Divide numerator and denominator by 𝒆𝒙 , we get
1 − 𝑒𝑥 (1 − 𝑒 𝑥 )Τ𝑒 𝑥
lim 𝑥
= lim
𝑥→∞ 1 + 2𝑒 𝑥→∞ (1 + 2𝑒 𝑥 )Τ𝑒 𝑥
1
𝑒𝑥
−1 0−1 1
= lim 1 = = − .
𝑥→∞ +2 0+2 2
𝑒𝑥
𝑒 3𝑥 − 𝑒−3𝑥
2. lim 3𝑥 −3𝑥
.
𝑥→∞ 𝑒 + 𝑒
Left to the student. The same strategy as the previous one.
𝑒 3𝑥 − 𝑒 −3𝑥
3. lim 3𝑥 −3𝑥
.
𝑥→−∞ 𝑒 + 𝑒
Divide numerator and denominator by 𝒆−𝟑𝒙 , we get
𝑒 3𝑥 − 𝑒 −3𝑥 (𝑒 3𝑥 − 𝑒 −3𝑥 )Τ𝑒 −3𝑥
lim 3𝑥 −3𝑥
= lim
𝑥→−∞ 𝑒 + 𝑒 𝑥→−∞ (𝑒 3𝑥 + 𝑒 −3𝑥 )Τ𝑒 −3𝑥
𝑒 6𝑥 − 1 0−1
= lim 6𝑥 = = −1 .
𝑥→−∞ 𝑒 + 1 0+1
Note that:
the line 𝑦 = −1 is a horizontal asymptote of the function
𝑒 3𝑥 − 𝑒 −3𝑥
𝑓 𝑥 = .
𝑒 3𝑥 + 𝑒 −3𝑥
4. lim [ ln 2 + 𝑥 − ln(1 + 𝑥) ] .
𝑥→∞
In this case we use the properties of ln 𝑥 as the following:
2+𝑥
lim [ ln 2 + 𝑥 − ln(1 + 𝑥) ]= lim ln
𝑥→∞ 𝑥→∞ 1+𝑥
𝑥
[Recall: ln = ln 𝑥 − ln 𝑦]
𝑦
2+𝑥
= ln lim
𝑥→∞ 1+𝑥
2
+1 0+1
𝑥
= ln lim 1 = ln( )
𝑥→∞ 𝑥+1 0+1
= ln(1) = 0 .
Example 5
Evaluate
lim sin 𝑥 .
𝑥→∞
Solution
As 𝑥 increases, the values of sin 𝑥 oscillate between 1 and
− 1 infinitely often and so they don’t approach any definite
number. Thus, lim sin 𝑥 does not exist.
𝑥→∞
Note that:
Similar reasoning shows that lim cos 𝑥 does not exist.
𝑥→∞
Infinite Limits at Infinity
Example: Evaluate
a) lim x 3
x →
3
b) lim x
x → −
Example 6
Find the horizontal asymptotes of the
following function:
𝑓 𝑥 = 5𝑥 3 − 4𝑥 2 + 8
Example 7
Prove that
Thank you for listening.
