Functions
Chapter 1
University of Namibia
Faculty of Engineering & IT
Department of Mechanical Engineering
March 4, 2024
Outline of Lecture
Introduction
Limits and Continuity
Exponential Functions
Logarithmic Functions
Hyperbolic Functions
Table of Contents
Introduction
Limits and Continuity
Exponential Functions
Logarithmic Functions
Hyperbolic Functions
Introduction
4.1 Introduction
Definition 4.1.1: Definition of a function
A function from a set A to a set B is a rule that assigns to each element x
in set A exactly one element y = f (x) as shown in Figure 1.
Figure 1: Definition of a function
1.
y = f (x) is called the image of x under f while x is called the
pre-image of y under f .
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Introduction
2.
The sets A and B are called Domain and Codomain of the function
f respectively; often denoted by Df and Cf .
3.
The set of all possible values of f as x varies throughout the domain is
called the range of the function f , often denoted by Rf and it is given
by
Rf = {f (x)|x ∈ A)}
(1)
Since y = f (x), then x is the independent variable and y is the dependent
variable.
Example: Consider the function given by
A(r ) = πr 2
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Introduction
Solution: The above function gives the area of a circle as a function of its
radius. The area of a circle of radius 2 is A(2) = 4π.
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Limit at a point
Table of Contents
1
Introduction
2
Limit at a point
3
Improper limits and continuity
4
Exponential Functions
5
Logarithmic Functions
6
Hyperbolic Functions
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Table of Contents
Introduction
Limits and Continuity
Exponential Functions
Logarithmic Functions
Hyperbolic Functions
Limit at a point
Limit at a point
To introduce the concept of limits let’s investigate the behavior of the
function defined by
x3 − 8
f (x) = 2
x −4
for values of near 2. This function is not defined at x = 2. The following
table gives values of f (x) for values of x close to 2, but not equal to 2.
x
f (x)
limx→2 f (x)
1.9
2.926
1.99
2.993
1.999
2.999
−→
2
*
*
2.001
3.001
←−
2.01
3.008
2.1
3.076
Table 1: Guessing limx→2 f (x).
(UNAM)
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Limit at a point
From Table 1 we see that when x is close to 2 (on either side of 2), f (x) is
3
close to 3. We express this by saying “the limit of the function f (x) = xx 2 −8
−4
as x approaches 2 is equal to 3.” Denoted by
x3 − 8
=3
x→2 x 2 − 4
lim f (x) = lim
x→2
Definition 4.8.1 Definition of limit
We write
lim f (x) = L
x→a
(2)
and say that “the limit of f (x), as x approaches a, equals to L.
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Limit at a point
Figure 2 shows the graphs of the three functions where in each case,
limx→a f (x) = L.
Figure 2: limx→a f (x) = L in all three cases
From Figure 2 we observe that in (a), f (a) = L, in (b), f (a) ̸= L and in (c),
f (a) is not defined. However, in each case, it is true that limx→a f (x) = L.
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Limit at a point
Example: Guess the values of the following limits
a)
b)
c)
limx→1
x−1
x 2 −1
limx→1 g (x) where g (x) =
limt→0
√
(
x−1
,
x 2 −1
2,
if x ̸= 1
if x = 1
t 2 +9−3
t2
Solution:
a)
The function f (x) = xx−1
2 −1 is not defined when x = 1, we will use the
table to compute values of f (x) for values of x close to 1, but not
equal to 1 as shown below.
x
f (x)
0.99
0.502513
0.999
0.500250
0.9999
0.500025
−→
1
*
*
1.0001
0.499975
←−
1.001
0.499750
1.01
0.497512
Table 2: Guessing limx→1 f (x)
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Limit at a point
From Table 2, we observe that f (x) → 0.5 from either side as x → 1. So,
we can make an educated guess that
x −1
= 0.5
x→1 x 2 − 1
lim
b)
The function g (x) resulted from the function f (x) in (a) by assigning
the value 2 when x = 1. So, the limit of f (x) and g (x) as x
approaches 1 are equal as shown in Figures 3 and 4
Figure 3
(UNAM)
Figure 4
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Limit at a point
c)
Table 3 below shows the list of values of the function f (t) =
for several values of t near 0.
t
f (t)
-0.5
0.16662
-0.1
0.16662
-0.05
0.16666
0.01
0.16667
−→
0
*
*
0.01
0.16667
←−
0.05
0.16666
0.1
0.16662
√
t 2 +9−3
t2
0.5
0.16662
Table 3: Guessing limt→0 f (t)
From Table 3 as t approaches 0, the values of the function seem to
approach 0.1666666. . . and so we make an educated guess that
√
t2 + 9 − 3
1
lim
=
2
t→0
t
6
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Limit at a point
Definition 4.8.2 One-sided limits
A one-sided limit can be the left-hand limit which is the limit as f (x)
approaches a from the left written as
lim f (x) = L
x→a−
(3)
or the right-hand limit which is the limit as f (x) approaches a from the
right written as
lim+ f (x) = L
(4)
x→a
Figure 5: limx→a− f (x) = L
(UNAM)
Figure 6: limx→a+ f (x) = L
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Limit at a point
Definition 5.8.3 Existence of Limits
For the function f (x) the limit exists and it is given by
lim f (x) = L
x→a
if and only if
lim f (x) = lim+ f (x) = L
x→a−
x→a
Example: Find limx→1 f (x) for the function
(
1 − 2x, x < 1
f (x) =
x − 3, x ≥ 1
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Limit at a point
Solution:The graph of f (x) is shown in Figure 7 below.
Figure 7: Graph of f (x)
We have that
lim f (x) = lim (1 − 2x) = 1 − 2(1) = −1
x→1−
x→1
and
lim f (x) = lim (x − 3) = 1 − 3 = −2
x→1+
(UNAM)
x→1
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Limit at a point
Since
lim f (x) = −1 ̸= −2 = lim+ f (x)
x→1−
x→1
then limx→1 f (x) does not exist.
Theorem 4.8.1 Limit Laws
Let f (x) and g (x) be functions such that
lim f (x) = L1
x→a
and
lim g (x) = L2
x→a
and c be a constant. Then
1
limx→a [f (x) + g (x)] = limx→a f (x) + limx→a g (x) = L1 + L2
2
limx→a [f (x) − g (x)] = limx→a f (x) − limx→a g (x) = L1 − L2
3
limx→a [cf (x)] = c limx→a f (x) = cL1
4
limx→a [f (x)g (x)] = limx→a f (x) · limx→a g (x) = L1 L2
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Limit at a point
f (x)
g (x)
if limx→a g (x) = L2 ̸= 0
7
limx→a [f (x)]n = [limx→a f (x)]n , where n is a positive integer
8
limx→a c = c
9
limx→a x = a
10
limx→a x n = an , where n is a positive integer
√
√
limx→a n x = n a, where n is a positive integer (If n is even, we
assume that a > 0).
p
p
limx→a n f (x) = n limx→a f (x), where n is a positive integer (If n is
even, we assume that limx→a f (x) > 0).
12
(UNAM)
=
L1
L2 ,
limx→a
11
=
limx→a f (x)
limx→a g (x)
6
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Limit at a point
Example: Find the following limits
a)
limx→2 x 2
b)
limx→3 (x 2 + 5)
c)
limx→3 4x 2
d)
e)
limx→5 (2x 2 − 3x + 4)
limx→−2
x 3 +2x 2 −1
5−3x
Solution:
a)
b)
c)
limx→2 x 2 = limx→2 c · c = limx→2 · limx→2 x = 2 · 2 = 4
limx→3 (x 2 + 5) = limx→3 x 2 + limx→3 5 = 32 + 5 = 14
limx→3 4x 2 = 4 limx→3 x 2 = 4 · 32 = 36
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Limit at a point
d)
lim (2x 2 − 3x + 4) = lim (2x 2 ) − lim (3x) + lim 4
x→5
x→5
x→5
x→5
2
= 2 lim x − 3 lim x + lim 4
x→5
2
x→5
x→5
= 2(5 ) − 3(5) + 4
= 39
e)
x 3 + 2x 2 − 1
limx→−2 (x 3 + 2x 2 − 1)
=
x→−2
5 − 3x
limx→−2 (5 − 3x)
limx→−2 x 3 + 2 limx→−2 x 2 − limx→−2 1
=
limx→−2 5 − 3 limx→−2 x
3
(−2) + 2(−2)2 − 1
=
5 − 3(−2)
1
=−
11
lim
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Limit at a point
Theorem 4.8.2
If f (x) = an x n + an−1 x n−1 an−2 x n−2 + · · · + a0 is a polynomial function,
and c is any number, then
limx→c f (x) = f (c) = an c n + an−1 c n−1 + an−2 c n−2 + · · · + a0 .
Theorem 4.8.3
If f (x) and g (x) are polynomials, and c is a constant, then
lim
x→c
f (x)
f (c)
=
,
g (x)
g (c)
provided that g (c) ̸= 0
Example: Determine the following limits
a)
limx→3 x 2 (2 − x)
x 3 +4x 2 −3
x 2 +5
b)
limx→2
c)
limx→−5
x 2 −25
3(x+5)
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Limit at a point
Solution:
a)
Since f (x) = x 2 (2 − x) = −x 3 + 2x 2 is a polynomial, then
lim f (x) = f (3) = −(3)3 + 2(32 ) = −9
x→3
b)
We have that f (x) = x 3 + 4x 2 − 3 and g (x) = x 2 + 5. So,
f (2) = 23 + 4(22 ) − 3 = 21 and g (2) = 22 + 5 = 9. Thus,
f (x)
f (2)
21
7
=
=
=
x→2 g (x)
g (2)
9
3
lim
c)
We have that f (x) = x 2 − 25 and g (x) = 3(x + 5). So
f (−5) = (−5)2 − 25 and g (−5) = 3(−5 + 5) = 0, then
undefined. Hence
f (−5)
g (−5)
is
f (x)
(x − 5)
(x
+5)
x −5
−10
= lim
= lim
=
x→−5 g (x)
x→−5
x→−5
3
(x
+5)
3
3
lim
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Limit at a point
4.8.1 Trigonometric Functions and Square Roots
Consider the function
f (x) =
To guess
sin x
x
sin x
x→0 x
lim f (x) = lim
x→0
We consider the following table
x
f (x)
-0.01
0.9999833
-0.005
0.9999958
-0.001
0.9999998
−→
0
*
*
0.001
0.9999998
←−
0.05
0.9999958
0.01
0.9999833
Table 4: Guessing limx→0 f (x)
From Table 3 as x approaches 0, the values of the function seem to
approach 1. Therefore
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Limit at a point
sin x
=1
x→0 x
lim
(5)
We use Equation 5 to find the limit as x approaches 0 of trigonometric
functions of similar form.
Example: Determine the following limits
sin(3x)
4x
a)
limx→0
b)
limx→0 2x cot x
c)
limx→0
sin θ
2θ
d)
limx→0
3x−tan(2x)
2x
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Limit at a point
Solution:
a)
Since x → 0, then 3x → 0. Thus
sin(3x)
sin(3x)
= lim
x→0
x→0 4x · 3
4x
3
lim
3 sin(3x)
x→0 4
3x
3
sin(3x)
= lim
4 x→0 3x
3
= ·1
4
3
=
4
= lim
b)
Note that cot x =
(UNAM)
cos x
sin x
and 2x → 0 as x → 0. Thus
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Limit at a point
cos x
x→0
sin x
x
= 2 lim
· cos x
x→0 sin x
1
= 2 lim sin x · cos x
lim 2x cot x = lim 2x ·
x→0
x→0
x
limx→0 1
=2
· lim cos x
limx→0 sinx x x→0
1
=2· ·1
1
=2
c)
limθ→0
sin θ
2θ
(UNAM)
= limθ→0
sin θ
θ
sin 2θ
θ
. Thus
26 / 77
Limit at a point
sin θ
θ
θ→0 sin 2θ
θ
lim
=
=
=
=
∴ limθ→0
d)
sin θ
2θ
limx→0
=
sin θ
θ
sin 2θ
2 limθ→0 2θ
limθ→0
1
2
1
2
3x−tan(2x)
2x
(UNAM)
sin θ
θ
limθ→0 sinθ2θ
limθ→0 sinθ θ
2θ
limθ→0 sin
θ· 22
limθ→0
= limx→0
3
2
−
sin 2x
2x cos 2x
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Limit at a point
lim
x→0
3
2
−
3 sin 2x
sin 2x 1 = lim
−
·
x→0 2
2x cos 2x
2x
cos 2x
3
sin 2x
1
= lim − lim
· lim
x→0 2
x→0 2x
x→0 cos x
3
1
= −1·1=
2
2
√
2
in Example 4.14 (c) we made
Note that for the function f (x) = t t+9−3
2
an educated guess that
√
t2 + 9 − 3
1
lim
=
t→0
t2
6
However, this limit can also be computed algebraically as follows
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Limit at a point
lim
t→0
√
√
√
t2 + 9 − 3
t2 + 9 + 3
√
·
t2
t2 + 9 + 3
t2 + 9 − 9
√
= lim
t→0 t 2 ( t 2 + 9 + 3)
t2 + 9 − 3
= lim
t→0
t2
t2
√
t→0 t 2 ( t 2 + 9 + 3)
1
1
= lim √
=
2
t→0
6
t +9+3
= lim
Example: Compute the following limits
a)
limx→0
b)
limx→0
c)
√
√
√
2x+7− 7
x
3x+1−1
x
√
√
2+h− 2
limh→0
h
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Limit at a point
a)
√
lim
x→0
b)
lim
x→0
√
2x + 7 −
x
√
7
√ √
√
2x + 7 − 7
2x + 7 + 7
√
= lim
·√
x→0
x
2x + 7 + 7
2x + 7 − 7
√
= lim √
x→0 x( 2x + 7 + 7)
2
x
√
= lim √
x→3 x( 2x + 7 + 7)
2
1
√ =√
= lim √
x→0
2x + 7 + 7
7
√
3x + 1 − 1
3x + 1 + 1
·√
x
3x + 1 + 1
3x + 1 − 1
3
x
= lim √
= lim √
x→3 x→0 x( 3x + 1 + 1)
x( 3x + 1 + 1)
3
3
= lim √
=
x→0
2
3x + 1 + 1
3x + 1 − 1
= lim
x→0
x
(UNAM)
√
√
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Limit at a point
c)
lim
h→0
(UNAM)
√
2+h−
h
√
2
√
√
√
2+h− 2
2+h+ 2
√
= lim
·√
h→0
h
2x + 7 + 7
2+h−2
√
= lim √
h→0 h( 2 + h + 2)
h
√
= lim √
h→0 h
( 2 + h + 2)
2
√
= lim √
h→0
2+h+ 2
2
= √
2 2
1
=√
2
√
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Improper limits and continuity
4.8.2 Infinite Limits
Consider the function
f (x) =
1
x2
To find
1
x→0 x 2
lim f (x) = lim
x→0
Let’s consider the following table of values of f (x) for values of x close to
0, but not equal to 0.
x
f (x)
-0.05
400
-0.01
10,000
-0.001
1,000,000
−→
0
*
*
0.001
1,000,000
←−
0.01
10,000
0.05
400
Table 5: Guessing limx→0 f (x)
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Improper limits and continuity
As shown in Figure 8 and Table 5 the values of f (x) can be made
arbitrarily large by taking x values close enough to 0. The values of f (x)
do not approach a number, so limx→0 x12 does not exist.
Figure 8: limx→0
1
x2
To indicate this kind of behavior, we use the notation
lim
x→0
(UNAM)
1
=∞
x2
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Improper limits and continuity
Definition 4.8.4
Let f (x) be a function defined on both sides of a, except possibly at a
itself. Then
lim f (x) = ∞
(6)
x→a
This means the values of f (x) can be made arbitrarily large by taking x
sufficiently close to a, but not equal to a.
Definition 4.8.5
Let f (x) be a function defined on both sides of a, except possibly at a
itself. Then
lim f (x) = −∞
(7)
x→a
This means the values of f (x) can be made arbitrarily large negative by
taking x sufficiently close to a, but not equal to a.
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Improper limits and continuity
Similar definitions can be given for the one-sided infinite limits
lim f (x) = ∞,
x→a−
lim f (x) = −∞,
x→a−
lim f (x) = ∞
x→a+
lim f (x) = −∞
x→a+
as demonstrated on Figure 9 below
Figure 9: One-sided infinite limits
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Improper limits and continuity
Definition 4.8.6
The line x = a is called a vertical asymptote of the curve y = f (x) if at
least one of the following statements is true:
lim f (x) = ∞,
x→a
lim f (x) = ∞,
x→a−
lim f (x) = −∞ lim f (x) = −∞,
x→a
x→a−
lim f (x) = ∞
x→a+
lim f (x) = −∞
x→a+
Example: Determine the following limits
a)
f (x) =
i)
ii)
b)
limx→0− f (x)
limx→0+ f (x)
g (x) =
i)
ii)
1
x
2x
x−3
limx→3− g (x)
limx→3+ g (x)
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Improper limits and continuity
Solution
a)
i)
f (x) = x1 increases in a negative direction as x approaches 0 from the
left. Thus
lim− f (x) = −∞
x→0
ii)
f (x) =
1
x
increases as x approaches 0 from the right. Thus
lim f (x) = +∞
x→0−
b)
i)
2x
increases in a negative direction as x approaches 3 from
g (x) = x−3
the left. Thus
lim− f (x) = −∞
x→3
ii)
g (x) =
2x
x−3
increases as x approaches 3 from the right. Thus
lim f (x) = +∞
x→3+
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Improper limits and continuity
4.8.3 Limits to Infinity
In this section, we want to look at the limit of the function f (x) as x → ∞
or x → −∞.
Consider the function
f (x) =
1
x
As shown in Figure 10 below.
Figure 10: limx→±∞
(UNAM)
1
x
39 / 77
Improper limits and continuity
This function is defined for all real numbers except at x = 0. Thus
a)
b)
When x → ∞,
1
x
When x → −∞,
→ 0+ .
1
x
→ 0− .
NB: If f (x) is a constant function. i.e. f (x) = k, then
limx→∞ f (x) = k = limx→−∞ f (x)
Theorem 4.8.4 Laws of limits to infinity
Let f (x) and g (x) be functions such that
lim f (x) = L1
x→∞
and
lim g (x) = L2 ,
x→∞
where L1 , L2 ∈ R
and c be a constant. Then
(UNAM)
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Improper limits and continuity
1.
limx→∞ [f (x) + g (x)] = limx→∞ f (x) + limx→∞ g (x) = L1 + L2
2.
limx→∞ [f (x) − g (x)] = limx→∞ f (x) − limx→∞ g (x) = L1 − L2
3.
4.
5.
limx→∞ [cf (x)] = c limx→∞ f (x) = cL1
limx→∞ [f (x)g (x)] = limx→∞ f (x) · limx→∞ g (x) = L1 L2
limx→∞
f (x)
g (x)
=
limx→∞ f (x)
limx→∞ g (x)
=
L1
L2 ,
if limx→∞ g (x) = L2 ̸= 0
Example: Determine the following limits
a)
limx→∞ 5 + x1
b)
limx→−∞
√
π 3
x2
Solution
a)
b)
limx→∞ 5 + x1 = limx→∞ 5 + limx→∞ x1 = 5 + 0 = 5.
√
√
√
limx→−∞ πx 23 = π 3 limx→−∞ x12 = π 3 · 0 = 0.
(UNAM)
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Improper limits and continuity
4.8.4 Limits of Rational Functions as x → ±∞
Consider the function
r (x) =
P(x)
Q(x)
To find
lim r (x)
x→±∞
Divide P(x) and Q(x) by the leading (highest) power of x in Q(x).
Here are three cases for limits of rational functions as x → ±∞:
(1.)
If deg (P(x)) < deg (Q(x)), limx→±∞ r (x) = 0.
(2.)
If deg (P(x)) = deg (Q(x)), limx→±∞ r (x) = bann , where an and bn are
the leading coefficients of P(x) and Q(x) respectively.
(3.)
If deg (P(x)) > deg (Q(x)), limx→±∞ r (x) = ±∞.
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Improper limits and continuity
Example: Determine the following limits
a)
b)
c)
d)
limx→∞ 11x+2
2x 3 −1
15x
limx→−∞ − 7x+4
5x 2 +8x−3
3x 2 +2
−4x 3 +7x
limx→∞ 2x 2 −3x−10
limx→∞
Solution
a)
P(x) = 11x + 2 and Q(x) = 2x 3 − 1. Thus
deg (P(x)) = 1 < 3 = deg (Q(x)). So,
11
+ x23
11x + 2
11 2 1
x2
=
lim
,
, ,
→ 0 as x → ∞
x→∞ 2 − 13
x→∞ 2x 3 − 1
x2 x3 x3
x
0
= =0
2
P(x) = 15x and Q(x) = 7x + 4. Thus deg (P(x)) = 1 = deg (Q(x)).
So,
lim
b)
(UNAM)
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Improper limits and continuity
lim −
x→−∞
c)
15
15x
= − lim
,
x→−∞ 7 + 4
7x + 4
x
15
=−
7
P(x) = 5x 2 + 8x − 3 and Q(x) = 3x 2 + 2. Thus,
deg (P(x)) = 2 = deg (Q(x)). So,
5 + x8 − x32
5x 2 + 8x − 3
=
lim
,
x→∞
x→∞
3x 2 + 2
3 + x22
5
=
3
lim
d)
4
→ 0 as x → −∞
x
8 3 2
, ,
→ 0 as x → ∞
x x2 x2
P(x) = −4x 3 + 7x and Q(x) = 2x 2 − 3x − 10. Thus
deg (P(x)) = 3 > 2 = deg (Q(x)). So,
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Improper limits and continuity
−4x + x7
−4x 3 + 7x
=
lim
x→∞ 2 − 3 − 102
x→∞ 2x 2 − 3x − 10
x
x
−4x
= lim
x→∞ 2
= −2 lim x
lim
x→∞
= −2 · ∞ = −∞
Theorem 4.8.5 The Sandwich (Squeeze) Theorem
Let f (x), g (x) and h(x) be functions such that f (x) ≤ h(x) ≤ g (x) for all
x in the interval containing a (except possibly at a). If
lim f (x) = L = lim g (x)
x→a
x→a
then
lim h(x) = L
x→a
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Improper limits and continuity
Example: Determine the following limits
a)
limx→0 x 2 sin x12
b)
limx→−∞ 2 + sinx x
Solution
a)
Since
−1 ≤ sin x ≤ 1
then
−1 ≤ sin
Thus,
1
≤1
x2
−x 2 ≤ x 2 sin
and
1
≤ x2
x2
lim (−x 2 ) = 0 = lim x 2
x→0
(UNAM)
x→0
46 / 77
Improper limits and continuity
Therefore,
lim x 2 sin
x→0
b)
We have that
lim
x→−∞
Since
then
2+
sin x
sin x sin x
= 2 + lim
= lim 2 + lim
x→−∞ x
x→−∞
x→−∞ x
x
−1 ≤ sin x ≤ 1
−
So,
1
sin x
1
≤
≤
x
x
x
lim −
x→−∞
(UNAM)
1
=0
x2
1
1
= 0 = lim
x→−∞ x
x
47 / 77
Improper limits and continuity
Therefore
sin x
=0
x→−∞ x
lim
∴ limx→−∞ 2 +
(UNAM)
sin x
x
=2+0=2
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Improper limits and continuity
4.8.5 Continuous Functions
Definition 4.8.7 Continuity at a point
A function y = f (x) is continuous at a point a on its domain if
lim f (x) = f (a)
x→a
(8)
Definition 4.8.7 implicitly requires three things if f (x) is continuous at a:
1.
f (a) is defined
2.
limx→a f (x) exists
3.
limx→a f (x) = f (a)
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Improper limits and continuity
Definition 4.8.8 Continuity at an end-point
A function is continuous from the right at x = a if
lim f (x) = f (a)
x→a+
(9)
and is continuous from the left at x = a if
lim f (x) = f (a)
x→a−
(10)
Discontinuity at a point
If a function f is not continuous at a point x = a, we say that f (x) is
discontinuous at x = a and we call x = a a point of discontinuity of f (x).
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Improper limits and continuity
Example
a)
Determine whether the function
(
1 − x 2, 0 < x ≤ 1
f (x) =
x, x > 1
is continuous at x = 1
b)
Show that the function f (x) = 1 −
interval [−1, 1].
√
1 − x 2 is continuous on the
Solution
a)
We have that
lim f (x) = lim (1 − x 2 ) = 1 − 12 = 0
x→1−
x→1
and
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Improper limits and continuity
lim f (x) = lim x = 1
x→1+
x→1
and
f (1) = 1 − (1)2 = 0
Since limx→1− f (x) = 0 ̸= 1 = limx→1+ f (x), then f (x) is not continuous
at x = 1.
b)
If −1 < a < 1, then
p
lim f (x) = lim 1 − 1 − x 2
x→a
x→a
p
= 1 − lim 1 − x 2
x→a
p
= 1 − 1 − a2 = f (a)
Therefore f (x) is continuous at x = a if −1 < a < 1.
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Improper limits and continuity
Furthermore,
x→−1
and
∴ f (x) = 1 −
√
(UNAM)
p
1 − 1 − x2
x→−1
p
= 1 − lim
1 − x2
x→−1
q
= 1 − 1 − (−1)2 = f (−1)
lim + f (x) = lim
p
lim f (x) = lim 1 − 1 − x 2
x→1
x→1−
p
= 1 − lim 1 − x 2
x→1
p
= 1 − 1 − 12 = f (1)
1 − x 2 is continuous on the interval [−1, 1].
53 / 77
Improper limits and continuity
Exercise 4.6
1.
Consider the function
(
1 − x 2 , x ̸= 1
f (x) =
2, x = 1
a)
b)
c)
2.
Graph the function f (x).
Find limx→1− f (x) and limx→1+ f (x).
Does limx→1 f (x) exist? If so, what is it. If not, why not?
Determine the following limits
a)
b)
c)
d)
e)
f)
limx→0 1+√33x+1
limx→0 sinx2x
5
4
+31
limx→±∞ 10x +x
6
√ x
2+ √x
limx→∞ 2− x
3
limx→2 x−2
limx→∞ f (x) if
(UNAM)
2x 2
x 2 +1
< f (x) <
2x 2 +5
x2
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Improper limits and continuity
4
Find the value of a for which the function
( 2
x −1
, x ̸= −1
f (x) = x+1
a, x = −1
is continuous.
(UNAM)
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Table of Contents
Introduction
Limits and Continuity
Exponential Functions
Logarithmic Functions
Hyperbolic Functions
Exponential Functions
Exponential and Logarithmic Functions
(1.) EXPONENTIAL FUNCTIONS
Definition 4.4.1
Consider a function of the form
f (x) = ax ,
a>0
(11)
such a function is called an exponential function.
Let us consider three cases for the base a as follows:
(1.) If a = 1, then f (x) = 1x = 1.
(2.) If 0 < a < 1, then the graph of f (x) = ax has the following properties:
I.
II.
III.
f (x) does not intersect the x−axis however, it intersect the y −axis at
y = f (0) = 1.
f (x) > 0 for all x ∈ R. Thus, Df = R and Rf = (0. + ∞).
As x → +∞, f (x) = ax → 0 and as x → −∞, f (x) → +∞.
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Exponential Functions
Figure 11: Graph of f (x) =
For example, let a =
above.
(3.)
1
2,
the graph of f (x) =
x
1
2
x
1
2
is shown in Figure 11
If a > 1, then the graph of f (x) = ax has the following properties:
I.
II.
III.
f (x) does not intersect the x−axis however, it intersect the y −axis at
y = f (0) = 1.
f (x) > 0 for all x ∈ R. Since a > 1 then ax > 0 for all x ∈ Df . Thus,
Df = R and Rf = (0. + ∞).
As x → +∞, f (x) = ax → +∞ and as x → −∞, f (x) → 0.
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Exponential Functions
Figure 12: Graph of f (x) =
x
1
2
For example, let a = 2, the graph of f (x) = 2x is shown in Figure 12 above.
x
In general, f (x) = 1a = a−x is a reflection of g (x) = ax on the y −axis.
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Exponential Functions
Figure 13: Graph of f (x) = e x and g (x) = e −x
A particular example of an exponential function is when
a = e = 2.718 . . . .
The function f (x) = e x is called the natural exponential function.
Since e > 1 and e1 < 1, we can sketch the graphs of f (x) = e x and
x
g (x) = e1 = e −x on the same axis as shown above.
(UNAM)
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Table of Contents
Introduction
Limits and Continuity
Exponential Functions
Logarithmic Functions
Hyperbolic Functions
Logarithmic Functions
(2.) LOGARITHMIC FUNCTIONS
Definition 4.4.2
A logarithmic function is a function of the form
f (x) = loga x
(12)
We consider the following cases for the base a
(1.)
If 0 < a < 1, then the graph of f (x) = loga x has the following
properties:
I.
f (x) does not intersect the y −axis however, it intersect the x−axis at
x = 1. Thus, f (x) = loga x = 0 =⇒ x = a0 = 1.
II.
f (x) = loga x is only defined for x > 0. Thus Df = (0, +∞) and
Rf = R.
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Logarithmic Functions
For example, let a = 12 , the graph of f (x) = log 1 x is shown below.
2
Figure 14: Graph of f (x) = log 12 x
2.
If a > 1, then the graph of f (x) = loga x has the following properties:
I.
f (x) does not intersect the y −axis however, it intersect the x−axis at
x = 1. Thus, f (x) = loga x = 0 =⇒ x = a0 = 1.
II.
f (x) = loga x is only defined for x > 0. Thus Df = (0, +∞) and
Rf = R.
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Logarithmic Functions
For example, let a = 2, the graph of f (x) = log2 x is shown below.
Figure 15: Graph of f (x) = log2 x
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Logarithmic Functions
(3.) RELATIONSHIP BETWEEN
EXPONENTIAL & LOGARITHMIC FUNCTIONS
Let’s investigate the relationship between the exponential function
f (x) = e x and logarithmic functions g (x) = loge x = ln x by looking at
their graphs a shown in Figure ?? below.
Figure 16: Graph of f (x) = e x and g (x) = loge x = ln x
(UNAM)
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Logarithmic Functions
x
and the logarithmic
Similarly, for the exponential function f (x) = 12
function g (x) = log 1 x their graphs are shown in Figure 17 below.
2
Figure 17: Graph of f (x) =
(UNAM)
x
1
2
and g (x) = log 12 x
66 / 77
Logarithmic Functions
Observations: From Figures ?? and 17 we observe that:
logarithmic g (x) = loga x is a reflection of the exponential function
f (x) = ax on the line y = x.
The logarithmic function g (x) = loge x is the inverse of the
exponential function f (x) = e x i.e. g (x) = loge x = f −1 (x).
x
Similarly, g (x) = log 1 x is the inverse of f (x) = 12
i.e.
2
g (x) = log 1 x = f −1 (x).
2
(UNAM)
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Logarithmic Functions
Exercise 4.2
Sketch the graphs of the following functions of the same axis
(a)
(b)
(c)
(d)
f (x) = 3x
x
g (x) = 13
h(x) = log3 x
i(x) = log 1 x
3
(UNAM)
68 / 77
Table of Contents
Introduction
Limits and Continuity
Exponential Functions
Logarithmic Functions
Hyperbolic Functions
Hyperbolic Functions
Hyperbolic Functions
Definition 4.5.1
Hyperbolic functions, sinh x, cosh x, tanh x, etc. are certain even and odd
combinations of exponential functions e x and e −x .
The six main hyperbolic functions are given below:
e x − e −x
e x + e −x
sinh x
,
cosh x =
,
tanh x =
2
2
cosh x
1
2
1
2
csch x =
= x
,
sech x =
= x
sinh x
e − e −x
cosh x
e + e −x
x
−x
1
cosh x
e +e
coth x =
=
= x
tanh x
sinh x
e − e −x
sinh x =
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Hyperbolic Functions
The notation of hyperbolic functions implies a close relation to
trigonometric, sin x, cos x, tan x, etc. However, the relationship is algebraic
rather than geometric. Figures 18, 19 and 20 shows the graphs of sinh x,
cosh x and tanh x.
Figure 18: sinh x
Figure 19: cosh x
Figure 20: tanh
The graphs of hyperbolic sine and cosine functions can be sketched using
graphical addition as shown in Figures 18 and 19.
(UNAM)
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Hyperbolic Functions
Hyperbolic Identities
Similar trigonometric functions, hyperbolic functions satisfy a number of
identities as shown below:
(1.)
(2.)
(3.)
(4.)
sinh(−x) = − sinh(x)
sinh x is odd
cosh(−x) = cosh(x)
cosh x is even
tanh(−x) = − tanh(x)
tanh x is odd
cosh2 x − sinh2 x = 1
(5.) 1 − tanh2 x = sech2 x
(6.)
(7.)
(8.)
(9.)
coth2 x − 1 = csch2 x
sinh(x ± y ) = sinh x cosh y ± sinh y cosh x
cosh(x ± y ) = cosh x cosh y ± sinh x sinh y
tanh x ± tanh y
tanh(x ± y ) =
1 ± tanh x tanh y
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Hyperbolic Functions
Example: Solve the following exercises
(1.)
(2.)
Find the numerical value of each expression
i)
sinh(ln 2)
ii)
sech(0)
iii)
sinh(1)
Prove that
i)
ii)
(3.)
sinh(−x) = − sinh(x)
cosh x + sinh x = e x
Prove that tanh(x + y ) =
(UNAM)
tanh x+tanh y
1+tanh x tanh y
73 / 77
Hyperbolic Functions
Solution:
(1.)
(2.)
e ln 2 −e − ln 2
2
i)
sinh(ln 2) =
ii)
sech (0) =
iii)
sinh(1) =
i)
sinh(−x) =
ii)
cosh x + sinh x =
(UNAM)
=
2− 12
2
1
cosh(0)
=
2
e 0 +e −0
e 1 −e −1
2
=
e 2 −1
2e
e −x −e x
2
= −e
e x +e −x
2
x
3
4
=1
−e −x
2
+
=
= − sinh x
e x −e −x
2
=
2e x
2
= ex
74 / 77
Hyperbolic Functions
3.
tanh(x + y ) =
sinh(x+y )
cosh(x+y )
sinh(x + y )
sinh x cosh y + cosh x sinh y
=
cosh(x + y )
cosh x cosh y + sinh x sinh y
∴ tanh(x + y ) =
(UNAM)
=
sinh x cosh y
cosh x cosh y
sinh x cosh y
cosh x cosh y
=
tanh x + tanh y
1 + tanh x tanh y
+
+
sinh y cosh x
cosh x cosh y
sinh x sinh y
cosh x cosh y
tanh x+tanh y
1+tanh x tanh y .
75 / 77
Hyperbolic Functions
Exercise 4.3
Show that
x 2 −1
x 2 +1
a)
tanh(ln x) =
b)
(cosh x + sinh x)n = cosh nx + sinh nx
(UNAM)
76 / 77
The End of Chapter 1