MAT 060/051
Analytic Geometry and
Calculus 1
Luzviminda T. Ranara, PhD
Mathematics Department
Mindanao State University Main Campus
Marawi City
luzviminda.ranara@msumain.edu.ph
September 15, 2020
Limits and Continuity of Functions
- At the end of this chapter, the student is expected to:
Limits and Continuity of Functions
- At the end of this chapter, the student is expected to:
1. Define intuitively and precisely the limit of a function.
Limits and Continuity of Functions
- At the end of this chapter, the student is expected to:
1. Define intuitively and precisely the limit of a function.
2. Use different analytic techniques such as factoring and rationalization and the
properties of limits such as the algebra of limits and two special trigonometric
limits to evaluate limits of functions.
Limits and Continuity of Functions
- At the end of this chapter, the student is expected to:
1. Define intuitively and precisely the limit of a function.
2. Use different analytic techniques such as factoring and rationalization and the
properties of limits such as the algebra of limits and two special trigonometric
limits to evaluate limits of functions.
3. Evaluate one-sided limits using similar properties.
Limits and Continuity of Functions
- At the end of this chapter, the student is expected to:
1. Define intuitively and precisely the limit of a function.
2. Use different analytic techniques such as factoring and rationalization and the
properties of limits such as the algebra of limits and two special trigonometric
limits to evaluate limits of functions.
3. Evaluate one-sided limits using similar properties.
4. Understand continuity by recognizing unbounded behavior of functions.
- The two broad areas of calculus known as differential and integral
calculus are built on the foundation concept of a limits.
- The two broad areas of calculus known as differential and integral
calculus are built on the foundation concept of a limits.
- In the first two sections, our approach to this important concept will be
both intuitive(informal definition)-understanding what a limit is using
algebraic functions (numerical and graphical examples), and
formal(evaluating limit in a more mathematical manner).
- The two broad areas of calculus known as differential and integral
calculus are built on the foundation concept of a limits.
- In the first two sections, our approach to this important concept will be
both intuitive(informal definition)-understanding what a limit is using
algebraic functions (numerical and graphical examples), and
formal(evaluating limit in a more mathematical manner).
- Finally, the approach in the last section will be analytical (using
algebraic methods to compute for the value of a limit of a function).
- The two broad areas of calculus known as differential and integral
calculus are built on the foundation concept of a limits.
- In the first two sections, our approach to this important concept will be
both intuitive(informal definition)-understanding what a limit is using
algebraic functions (numerical and graphical examples), and
formal(evaluating limit in a more mathematical manner).
- Finally, the approach in the last section will be analytical (using
algebraic methods to compute for the value of a limit of a function).
- This chapter also includes limits of trigonometric functions, which are
presented as theorems.
- The two broad areas of calculus known as differential and integral
calculus are built on the foundation concept of a limits.
- In the first two sections, our approach to this important concept will be
both intuitive(informal definition)-understanding what a limit is using
algebraic functions (numerical and graphical examples), and
formal(evaluating limit in a more mathematical manner).
- Finally, the approach in the last section will be analytical (using
algebraic methods to compute for the value of a limit of a function).
- This chapter also includes limits of trigonometric functions, which are
presented as theorems.
- As one of the uses or applications of limits, continuity of functions is one
of the sections included in this chapter.
Intuitive Idea of Limits
Let a be a real number and let f be a function of x given by y = f (x).
Intuitive Idea of Limits
Let a be a real number and let f be a function of x given by y = f (x).In
Calculus and its applications, we are often interested in the values f (x) of a
function f when x is very closed to a number a, but not necessarily equal to
a.
Intuitive Idea of Limits
Let a be a real number and let f be a function of x given by y = f (x).In
Calculus and its applications, we are often interested in the values f (x) of a
function f when x is very closed to a number a, but not necessarily equal to
a.
In many instances, as a matter of fact, a is not in the domain of f ; that is,
f (a) is undefined. Roughly speaking, we ask the following questions:
Intuitive Idea of Limits
Let a be a real number and let f be a function of x given by y = f (x).In
Calculus and its applications, we are often interested in the values f (x) of a
function f when x is very closed to a number a, but not necessarily equal to
a.
In many instances, as a matter of fact, a is not in the domain of f ; that is,
f (a) is undefined. Roughly speaking, we ask the following questions:
As x gets closer and closer to a (but x 6= a), does f (x) get closer and closer
to some number L?
If the answer is yes, we say that the limit of f (x), as x approaches a,
equals L or the limit of f (x), as x approaches a is L and we write in
symbol as either the following:
If the answer is yes, we say that the limit of f (x), as x approaches a,
equals L or the limit of f (x), as x approaches a is L and we write in
symbol as either the following:
1. f (x) −→ L, as x −→ a
If the answer is yes, we say that the limit of f (x), as x approaches a,
equals L or the limit of f (x), as x approaches a is L and we write in
symbol as either the following:
1. f (x) −→ L, as x −→ a
2. lim f (x) = L
x→a
If the answer is yes, we say that the limit of f (x), as x approaches a,
equals L or the limit of f (x), as x approaches a is L and we write in
symbol as either the following:
1. f (x) −→ L, as x −→ a
2. lim f (x) = L
x→a
To Illustrate the idea above, let us consider the following examples:
If the answer is yes, we say that the limit of f (x), as x approaches a,
equals L or the limit of f (x), as x approaches a is L and we write in
symbol as either the following:
1. f (x) −→ L, as x −→ a
2. lim f (x) = L
x→a
To Illustrate the idea above, let us consider the following examples:
Example 1.1 Consider the values of the function f (x) = 12 (3x − 1) as x gets
very close to x = 4. Note that 4 ∈ df = R. That is, 4 is an element of the
domain of the function f . But then we are interested primarily in the
behavior of f (x) as x takes values closer and closer to a = 4 but not
neccessarily equal to 4.
If the answer is yes, we say that the limit of f (x), as x approaches a,
equals L or the limit of f (x), as x approaches a is L and we write in
symbol as either the following:
1. f (x) −→ L, as x −→ a
2. lim f (x) = L
x→a
To Illustrate the idea above, let us consider the following examples:
Example 1.1 Consider the values of the function f (x) = 12 (3x − 1) as x gets
very close to x = 4. Note that 4 ∈ df = R. That is, 4 is an element of the
domain of the function f . But then we are interested primarily in the
behavior of f (x) as x takes values closer and closer to a = 4 but not
neccessarily equal to 4.
So, to address this interest, we consider the following tables:
x
f (x) =
1
2 (3x
− 1)
3.99
5.485
3.999
5.4985
3.99999
5.49985
3.99999
5.499985
Table: The values of the function f (x) = 12 (3x − 1) for x < 4.
x
f (x) =
1
2 (3x
− 1)
3.99
5.485
3.999
5.4985
3.99999
5.49985
3.99999
5.499985
Table: The values of the function f (x) = 12 (3x − 1) for x < 4.
Considering the values of x < 4,observe that as the values of x get closer to
4, the values of f (x) become closer to 5.5.
x
f (x) 12 (3x
− 1)
4.01
5.515
4.001
5.5015
4.0001
5.50015
4.00001
5.500015
Table: The values of the function f (x) = 12 (3x − 1) for x > 4.
x
f (x) 12 (3x
− 1)
4.01
5.515
4.001
5.5015
4.0001
5.50015
4.00001
5.500015
Table: The values of the function f (x) = 12 (3x − 1) for x > 4.
Observe also that for x > 4, as the values of x get closer to 4, the values of
f (x) become closer to 5.5, also.
x
f (x) 12 (3x
− 1)
4.01
5.515
4.001
5.5015
4.0001
5.50015
4.00001
5.500015
Table: The values of the function f (x) = 12 (3x − 1) for x > 4.
Observe also that for x > 4, as the values of x get closer to 4, the values of
f (x) become closer to 5.5, also.
Thus, as x gets closer to 4, both from the left (that is, x < 4) and from the
right of 4 (that is, x > 4), the values of f (x) become closer to 5.5.
x
f (x) 12 (3x
− 1)
4.01
5.515
4.001
5.5015
4.0001
5.50015
4.00001
5.500015
Table: The values of the function f (x) = 12 (3x − 1) for x > 4.
Observe also that for x > 4, as the values of x get closer to 4, the values of
f (x) become closer to 5.5, also.
Thus, as x gets closer to 4, both from the left (that is, x < 4) and from the
right of 4 (that is, x > 4), the values of f (x) become closer to 5.5.
Hence, we say that the limit of f (x) = 12 (3x − 1) as x gets very close to
x = 4 is 5.5.
x
f (x) 12 (3x
− 1)
4.01
5.515
4.001
5.5015
4.0001
5.50015
4.00001
5.500015
Table: The values of the function f (x) = 12 (3x − 1) for x > 4.
Observe also that for x > 4, as the values of x get closer to 4, the values of
f (x) become closer to 5.5, also.
Thus, as x gets closer to 4, both from the left (that is, x < 4) and from the
right of 4 (that is, x > 4), the values of f (x) become closer to 5.5.
Hence, we say that the limit of f (x) = 12 (3x − 1) as x gets very close to
x = 4 is 5.5.
In symbol,
As x → 4, f (x) → 5.5 or
1
lim (3x − 1) = 5.5.
x→4 2
Note that from the preceeding example, the number a = 4 could actually
have been substituted for x in the given function f , thus obtaining L = 5.5
as the limit of the function, i.e.,
Note that from the preceeding example, the number a = 4 could actually
have been substituted for x in the given function f , thus obtaining L = 5.5
as the limit of the function, i.e.,
1
1
1
lim (3x − 1) = (3(4) − 1) = (11) = 5.5.
x→4 2
2
2
Note that from the preceeding example, the number a = 4 could actually
have been substituted for x in the given function f , thus obtaining L = 5.5
as the limit of the function, i.e.,
1
1
1
lim (3x − 1) = (3(4) − 1) = (11) = 5.5.
x→4 2
2
2
This method of evaluating limits is called direct substitution. But the
next examples show that it is not always possible to find the number L by
merely substituting a for x.
Example 1.2 Determine the values of f (x) =
to a = 1 or determine
x−1
.
x→1 x2 − 1
lim
x−1
as x takes values closer
x2 − 1
Example 1.2 Determine the values of f (x) =
to a = 1 or determine
x−1
.
x→1 x2 − 1
lim
Consider the following solutions:
x−1
as x takes values closer
x2 − 1
Example 1.2 Determine the values of f (x) =
to a = 1 or determine
x−1
as x takes values closer
x2 − 1
x−1
.
x→1 x2 − 1
lim
Consider the following solutions:
1. Constructing the table of values for x and f (x):
x
0.9
0.99
0.999
0.99999
0.99999
f (x) 0.52631 0.50251 0.50025 0.500025 0.5000025
Table: For x < 1
-
x
1.1
1.01
1.001
1.0001
1.00001
f (x) 0.4761 0.4975 0.49975 0.499975 0.4999975
Table: For x > 1
-
x
1.1
1.01
1.001
1.0001
1.00001
f (x) 0.4761 0.4975 0.49975 0.499975 0.4999975
Table: For x > 1
Note that the values of f (x) tend to get closer to .5 as x approaches to 1.
Thus,
x−1
lim 2
= .5
x→1 x − 1
- 2. Simplifying the function by factoring:
- 2. Simplifying the function by factoring:
Note that, by direct substitution, we get
- 2. Simplifying the function by factoring:
Note that, by direct substitution, we get
x−1
1−1
0
=
=
.
x→1 x2 − 1
12 − 1
0
lim
- 2. Simplifying the function by factoring:
Note that, by direct substitution, we get
x−1
1−1
0
=
=
.
x→1 x2 − 1
12 − 1
0
lim
Thus, we cannot use direct substitution. However, simplifying the
expression by factoring, we get
- 2. Simplifying the function by factoring:
Note that, by direct substitution, we get
x−1
1−1
0
=
=
.
x→1 x2 − 1
12 − 1
0
lim
Thus, we cannot use direct substitution. However, simplifying the
expression by factoring, we get
1
x−1
x−1
1
=
lim
=
lim
=
= 0.5,
x→1 x2 − 1
x→1 (x − 1)(x + 1)
x→1 x + 1
2
lim
where
x−1
1
=
since x 6= 1, x only approaches 1.
x2 − 1
x+1
x−4
, find lim f (x).
x→4
x−2
- Example 1.3 If f (x) = √
x−4
, find lim f (x).
x→4
x−2
- Solution 1: Construction of the table of values:
- Example 1.3 If f (x) = √
x−4
, find lim f (x).
x→4
x−2
- Solution 1: Construction of the table of values:
Constructing the table of values, we get
- Example 1.3 If f (x) = √
x−4
, find lim f (x).
x→4
x−2
- Solution 1: Construction of the table of values:
Constructing the table of values, we get
x
3.9
3.99
3.999 4 4.001
4.01
4.1
f (x) 3.9748 3.9975 3.9997
4.0002 4.0025 4.0248
- Example 1.3 If f (x) = √
Table: For x → 4
x−4
, find lim f (x).
x→4
x−2
- Solution 1: Construction of the table of values:
Constructing the table of values, we get
x
3.9
3.99
3.999 4 4.001
4.01
4.1
f (x) 3.9748 3.9975 3.9997
4.0002 4.0025 4.0248
- Example 1.3 If f (x) = √
Table: For x → 4
- Note that the values of f (x) tend to get closer to 4 as x approaches to 4.
Thus,
x−4
lim √
= 4.
x→4
x−2
- Simplifying the expression by rationalizing:
- Simplifying the expression by rationalizing:
First, we use direct substitution to obtain
- Simplifying the expression by rationalizing:
First, we use direct substitution to obtain
4−4
0
x−4
lim √
=√
= ,
x→4
0
x−2
4−2
- Simplifying the expression by rationalizing:
First, we use direct substitution to obtain
4−4
0
x−4
lim √
=√
= ,
x→4
0
x−2
4−2
which is again an indeterminate form. However rationalizing the
denominator of the function, we get
- Simplifying the expression by rationalizing:
First, we use direct substitution to obtain
4−4
0
x−4
lim √
=√
= ,
x→4
0
x−2
4−2
which is again an indeterminate form. However rationalizing the
denominator of the function, we get
√
x−4
x−4
x+2
lim √
= lim √
·√
x→4
x→4
x−2
x−2
x+2
- Simplifying the expression by rationalizing:
First, we use direct substitution to obtain
4−4
0
x−4
lim √
=√
= ,
x→4
0
x−2
4−2
which is again an indeterminate form. However rationalizing the
denominator of the function, we get
√
x−4
x−4
x+2
lim √
= lim √
·√
x→4
x→4
x−2
x−2
x+2
√
(x − 4) x + 2
= lim
x→4
x−4
= lim
x→4
√
x+2
√
= lim x + 2
x→4
√
=
2 + 2 = 4.
√
= lim x + 2
x→4
√
=
2 + 2 = 4.
Thus,
x−4
lim √
= 4.
x→4
x−2
- Each solution of the preceeding two examples shows that the following
algebraic manipulations enumerated below can be used to simplify the
task of finding limits especially when the result of direct substitution is
:
0
0
- Each solution of the preceeding two examples shows that the following
algebraic manipulations enumerated below can be used to simplify the
task of finding limits especially when the result of direct substitution is
:
1. simplifying rational expressions: just as used in Example 1.2,
0
0
- Each solution of the preceeding two examples shows that the following
algebraic manipulations enumerated below can be used to simplify the
task of finding limits especially when the result of direct substitution is
:
1. simplifying rational expressions: just as used in Example 1.2,
2. rationalizing denominator or numerator: just as used in Example 1.3.
0
0
The following example shows that in some instances, the limit does not exist:
The following example shows that in some instances, the limit does not exist:
1
Example 1.4 Evalute the limit of f (x) = as x approaches to a = 0.
x
The following example shows that in some instances, the limit does not exist:
1
Example 1.4 Evalute the limit of f (x) = as x approaches to a = 0.
x
Solution: Consider the following tables :
The following example shows that in some instances, the limit does not exist:
1
Example 1.4 Evalute the limit of f (x) = as x approaches to a = 0.
x
Solution: Consider the following tables :
x 1 0.1 0.01 0.001 0.0001 0.00001
f (x) 1 10 100 1000 10000 100000
Table: For x > 0
The following example shows that in some instances, the limit does not exist:
1
Example 1.4 Evalute the limit of f (x) = as x approaches to a = 0.
x
Solution: Consider the following tables :
x 1 0.1 0.01 0.001 0.0001 0.00001
f (x) 1 10 100 1000 10000 100000
Table: For x > 0
x
-1 −0.1 −0.01 −0.001 −0.0001 −0.00001
f (x) −1 −10 −100 −10000 −10000 −100000
Table: For x < 0
1
does not exist.
x→0 x
- In the table, it can be observed that lim
1
does not exist.
x→0 x
- Exercise 1.5 Use intuitive manner(that is, constructing table of values) to
find each of the following limits, if it exists,
- In the table, it can be observed that lim
1
x→−2 x + 2
2. lim (−4)
x→2 √
3. lim 102 − 5x
1.
lim
x→0
4.
lim
√ x
x→ 2
2
−3
5. lim [(2x − 7) + 4 − 3x]
x→1
x2 − 4
x→−2 x + 2
r2 + 2r − 3
7. lim 2
r→−3 r + 7r + 12
s2 − 16
8. lim √
s→4
s−2
6.
lim
ASSESSMENT TEST
NAME:
SIGNATURE:
SECTION:
DATE:
SCORE:
Exercise 1.6 Use intuitive manner and find each of the following limit, if it
exists.
1−t
1. lim
t→1 t − 1
2. lim 3x + 4
x→1
h3 − 8
3. lim 2
h→2 h − 4
Formal Definition of Limit
- The intention of the informal discussion in section 1.1 was to give you an
intuitive(unpremiditated) grasp of when a limit does or does not exist.
However, it is neither desirable nor practical to reach into a conclusion
on limit existence by merely basing on such intuition. We must be able
to evaluate a limit or discern its existence in a somewhat technical
fashion. The theorems that we shall consider and discuss in this section
establish such a means.
Formal Definition of Limit
- The intention of the informal discussion in section 1.1 was to give you an
intuitive(unpremiditated) grasp of when a limit does or does not exist.
However, it is neither desirable nor practical to reach into a conclusion
on limit existence by merely basing on such intuition. We must be able
to evaluate a limit or discern its existence in a somewhat technical
fashion. The theorems that we shall consider and discuss in this section
establish such a means.
- Let us consider the table of values of the function given in Example 1.1.
In detail, the variation of f (x) = 12 (3x − 1) when x is close to 4 can be
written in the following conditional statements:
- If 3.9 < x < 4.1, then 5.35 < f (x) < 5.65.
- If 3.9 < x < 4.1, then 5.35 < f (x) < 5.65.
If 3.99 < x < 4.01, then 5.485 < f (x) < 5.515.
- If 3.9 < x < 4.1, then 5.35 < f (x) < 5.65.
If 3.99 < x < 4.01, then 5.485 < f (x) < 5.515.
If 3.999 < x < 4.001, then 5.4985 < f (x) < 5.5015.
- If 3.9 < x < 4.1, then 5.35 < f (x) < 5.65.
If 3.99 < x < 4.01, then 5.485 < f (x) < 5.515.
If 3.999 < x < 4.001, then 5.4985 < f (x) < 5.5015.
If 3.9999 < x < 4.0001, then 5.49985 < f (x) < 5.50015.
- If 3.9 < x < 4.1, then 5.35 < f (x) < 5.65.
If 3.99 < x < 4.01,
If 3.999 < x < 4.001,
If 3.9999 < x < 4.0001,
If 3.99999 < x < 4.00001,
then
then
then
then
5.485 < f (x) < 5.515.
5.4985 < f (x) < 5.5015.
5.49985 < f (x) < 5.50015.
5.499985 < f (x) < 5.500015.
- If 3.9 < x < 4.1, then 5.35 < f (x) < 5.65.
If 3.99 < x < 4.01,
If 3.999 < x < 4.001,
If 3.9999 < x < 4.0001,
If 3.99999 < x < 4.00001,
5.485 < f (x) < 5.515.
5.4985 < f (x) < 5.5015.
5.49985 < f (x) < 5.50015.
5.499985 < f (x) < 5.500015.
- If lim f (x) = L, we say that if we let the Greek letters "Epsilon " and
x→a
δ "Delta" denote small positive real numbers, then each of the
statements above is of the form:
then
then
then
then
- If 3.9 < x < 4.1, then 5.35 < f (x) < 5.65.
If 3.99 < x < 4.01,
If 3.999 < x < 4.001,
If 3.9999 < x < 4.0001,
If 3.99999 < x < 4.00001,
5.485 < f (x) < 5.515.
5.4985 < f (x) < 5.5015.
5.49985 < f (x) < 5.50015.
5.499985 < f (x) < 5.500015.
- If lim f (x) = L, we say that if we let the Greek letters "Epsilon " and
x→a
δ "Delta" denote small positive real numbers, then each of the
statements above is of the form:
If 4 − δ < x < 4 + δ , then 5.5 − < f (x) < 5.5 + .
then
then
then
then
- For example, let δ = 0.1 and = 0.15 in the first statement, we have
If 3.9 < x < 4.1, then 5.35 < f (x) < 5.65;
- For example, let δ = 0.1 and = 0.15 in the first statement, we have
If 3.9 < x < 4.1, then 5.35 < f (x) < 5.65;
- Let δ = 0.01 and = 0.015 in the second statement; and δ = 0.001 and
= 0.0015 in the third statement and so on.
- For example, let δ = 0.1 and = 0.15 in the first statement, we have
If 3.9 < x < 4.1, then 5.35 < f (x) < 5.65;
- Let δ = 0.01 and = 0.015 in the second statement; and δ = 0.001 and
= 0.0015 in the third statement and so on.
- In interval form, we have
- For example, let δ = 0.1 and = 0.15 in the first statement, we have
If 3.9 < x < 4.1, then 5.35 < f (x) < 5.65;
- Let δ = 0.01 and = 0.015 in the second statement; and δ = 0.001 and
= 0.0015 in the third statement and so on.
- In interval form, we have
If x is in the open interval (4 − δ, 4 + δ), then f (x) is in the open
interval (5.5 − , 5.5 + ) or equivalently,
If |x − 4| < δ , then |f (x) − 5.5| < - For example, let δ = 0.1 and = 0.15 in the first statement, we have
If 3.9 < x < 4.1, then 5.35 < f (x) < 5.65;
- Let δ = 0.01 and = 0.015 in the second statement; and δ = 0.001 and
= 0.0015 in the third statement and so on.
- In interval form, we have
If x is in the open interval (4 − δ, 4 + δ), then f (x) is in the open
interval (5.5 − , 5.5 + ) or equivalently,
If |x − 4| < δ , then |f (x) − 5.5| < - Since and δ are positive real numbers, the statements above can be
written as
If 0 < |x − 4| < δ , then 0 < |f (x) − 5.5| < - Definition 1.7 Let f be a function which is defined for all x on the open
interval I containing a, except posssibly at a itself. The limit of f (x) as
x approaches to a is L, written
- Definition 1.7 Let f be a function which is defined for all x on the open
interval I containing a, except posssibly at a itself. The limit of f (x) as
x approaches to a is L, written
lim f (x) = L
x→a
- Definition 1.7 Let f be a function which is defined for all x on the open
interval I containing a, except posssibly at a itself. The limit of f (x) as
x approaches to a is L, written
lim f (x) = L
x→a
if for every > 0, however small, there exists a δ > 0 such that
- Definition 1.7 Let f be a function which is defined for all x on the open
interval I containing a, except posssibly at a itself. The limit of f (x) as
x approaches to a is L, written
lim f (x) = L
x→a
if for every > 0, however small, there exists a δ > 0 such that
|f (x) − L| < whenever 0 < |x − a| < δ .
- Definition 1.7 Let f be a function which is defined for all x on the open
interval I containing a, except posssibly at a itself. The limit of f (x) as
x approaches to a is L, written
lim f (x) = L
x→a
if for every > 0, however small, there exists a δ > 0 such that
|f (x) − L| < whenever 0 < |x − a| < δ .
- From Definition 1.7 , if
lim f (x) = L,
x→a
we say that
- Definition 1.7 Let f be a function which is defined for all x on the open
interval I containing a, except posssibly at a itself. The limit of f (x) as
x approaches to a is L, written
lim f (x) = L
x→a
if for every > 0, however small, there exists a δ > 0 such that
|f (x) − L| < whenever 0 < |x − a| < δ .
- From Definition 1.7 , if
lim f (x) = L,
x→a
we say that
limit of f (x), as x approaches a, is L.
- Since can be arbitrarily small, f (x) can be made arbitrarily close to L
by choosing x sufficiently close to a.
- Since can be arbitrarily small, f (x) can be made arbitrarily close to L
by choosing x sufficiently close to a.
- That is, if x is in the open interval (a − δ, a + δ), then f (x) is in the
open interval (L − , L + ).
- Since can be arbitrarily small, f (x) can be made arbitrarily close to L
by choosing x sufficiently close to a.
- That is, if x is in the open interval (a − δ, a + δ), then f (x) is in the
open interval (L − , L + ).
- The following examples will illustrate how to show that lim f (x) = L
x→a
using the formal definition of limit:
- Example 1.8 Prove that lim 12 (3x − 1) = 11
2 .
x→4
- Example 1.8 Prove that lim 12 (3x − 1) = 11
2 .
x→4
- Let f (x) = 12 (3x − 1), a = 4, and L = 11
2 . By Definition 1.7, we must
show that for every > 0, there exists a number δ > 0 such that if
0 < |x − 4| < δ , then 12 (3x − 1) − 11
2 < .
- Example 1.8 Prove that lim 12 (3x − 1) = 11
2 .
x→4
- Let f (x) = 12 (3x − 1), a = 4, and L = 11
2 . By Definition 1.7, we must
show that for every > 0, there exists a number δ > 0 such that if
0 < |x − 4| < δ , then 12 (3x − 1) − 11
2 < .
- A clue to the choice of δ can be found by examining the inequality
11
1
2 (3x − 1) − 2 < which involve . Then we have the following
equivalent inequalities:
-
1
11
(3x − 1) −
<
2
2
-
1
11
(3x − 1) −
<
2
2
1
|(3x − 1) − 11| < 2
-
1
11
(3x − 1) −
<
2
2
1
|(3x − 1) − 11| < 2
|3x − 12| < 2
-
1
11
(3x − 1) −
<
2
2
1
|(3x − 1) − 11| < 2
|3x − 12| < 2
3 |x − 4| < 2
-
1
11
(3x − 1) −
<
2
2
1
|(3x − 1) − 11| < 2
|3x − 12| < 2
3 |x − 4| < 2
2
|x − 4| < 3
-
1
11
(3x − 1) −
<
2
2
1
|(3x − 1) − 11| < 2
|3x − 12| < 2
3 |x − 4| < 2
2
|x − 4| < 3
- The final inequality gives us the needed clue. If we take δ = 23 , then for
> 0 there exist a δ > 0 such that |x − 4| < δ , then
> 0 there exist a δ > 0 such that |x − 4| < δ , then
11
1
(3x − 1) −
2
2
=
1
|(3x − 1) − 11|
2
> 0 there exist a δ > 0 such that |x − 4| < δ , then
11
1
(3x − 1) −
2
2
1
|(3x − 1) − 11|
2
1
=
|3x − 12|
2
=
> 0 there exist a δ > 0 such that |x − 4| < δ , then
11
1
(3x − 1) −
2
2
1
|(3x − 1) − 11|
2
1
=
|3x − 12|
2
3
=
|x − 4|
2
=
> 0 there exist a δ > 0 such that |x − 4| < δ , then
11
1
(3x − 1) −
2
2
1
|(3x − 1) − 11|
2
1
=
|3x − 12|
2
3
=
|x − 4|
2
3
< δ
2
=
> 0 there exist a δ > 0 such that |x − 4| < δ , then
11
1
(3x − 1) −
2
2
=
=
=
<
=
1
|(3x − 1) − 11|
2
1
|3x − 12|
2
3
|x − 4|
2
3
δ
2 3
2
2
3
> 0 there exist a δ > 0 such that |x − 4| < δ , then
11
1
(3x − 1) −
2
2
=
=
=
<
=
=
1
|(3x − 1) − 11|
2
1
|3x − 12|
2
3
|x − 4|
2
3
δ
2 3
2
2
3
.
> 0 there exist a δ > 0 such that |x − 4| < δ , then
11
1
(3x − 1) −
2
2
=
=
=
<
=
=
Therefore, lim 21 (3x − 1) =
x→4
11
2 .
1
|(3x − 1) − 11|
2
1
|3x − 12|
2
3
|x − 4|
2
3
δ
2 3
2
2
3
.
- Example 1.9 Prove that the lim (5x − 3) = 2.
x→1
- Example 1.9 Prove that the lim (5x − 3) = 2.
x→1
- Proof: Let > 0. We show that there exists a δ > 0 such that
|(5x − 3) − 2| < whenever 0 < |x − 1| < δ
- Example 1.9 Prove that the lim (5x − 3) = 2.
x→1
- Proof: Let > 0. We show that there exists a δ > 0 such that
|(5x − 3) − 2| < whenever 0 < |x − 1| < δ
- Now, |(5x − 3) − 2| = |5x − 5| = 5|x − 1|. Hence, we show that
|x − 1| <
whenever 0 < |x − 1| < δ
5
- Example 1.9 Prove that the lim (5x − 3) = 2.
x→1
- Proof: Let > 0. We show that there exists a δ > 0 such that
|(5x − 3) − 2| < whenever 0 < |x − 1| < δ
- Now, |(5x − 3) − 2| = |5x − 5| = 5|x − 1|. Hence, we show that
|x − 1| <
whenever 0 < |x − 1| < δ
5
- Choose δ = 5 . Then, for > 0 there exists δ > 0 such that if
0 < |x − 1| < δ , then
|(5x − 3) − 2| = 5|x − 1| < 5δ = 5
5
=
|(5x − 3) − 2| = 5|x − 1| < 5δ = 5
5
=
Hence,
|(5x − 3) − 2| < whenever 0 < |x − 1| < δ
|(5x − 3) − 2| = 5|x − 1| < 5δ = 5
5
=
Hence,
|(5x − 3) − 2| < whenever 0 < |x − 1| < δ
Thus, by definition, lim (5x − 3) = 2.
x→1
1
- Example 1.10 Show that lim
does not exist.
x→0 x
1
- Example 1.10 Show that lim
does not exist.
x→0 x
1
- Suppose that lim
exists. Then for some number L,
x→0 x
1
- Example 1.10 Show that lim
does not exist.
x→0 x
1
- Suppose that lim
exists. Then for some number L,
x→0 x
1
= L.
lim
x→0 x
1
- Example 1.10 Show that lim
does not exist.
x→0 x
1
- Suppose that lim
exists. Then for some number L,
x→0 x
1
= L.
lim
x→0 x
- Thus, for each > 0, it is possible to find an interval (0 − δ, 0 + δ)
containing 0 such that whenever 0 < |x − 0| < δ ,
|f (x) − L| =
1
− L < .
x
1
- Example 1.10 Show that lim
does not exist.
x→0 x
1
- Suppose that lim
exists. Then for some number L,
x→0 x
1
= L.
lim
x→0 x
- Thus, for each > 0, it is possible to find an interval (0 − δ, 0 + δ)
containing 0 such that whenever 0 < |x − 0| < δ ,
|f (x) − L| =
1
− L < .
x
- Now,
x−
1 − xL
1
|f (x) − L| =
−L =
= (−L)
x
x
x
1
L
=L
x−
x
1
L
.
- Now,
x−
1 − xL
1
|f (x) − L| =
−L =
= (−L)
x
x
x
1
L
=L
x−
x
1
L
.
1
- Note that for x 6= 0, L
6= 0, which is a contradiction. Hence, we can not
find a δ such that for every > 0, if 0 < |x − 0| < δ , then
1
x
− L < .
- Now,
x−
1 − xL
1
|f (x) − L| =
−L =
= (−L)
x
x
x
1
L
=L
x−
x
1
L
.
1
- Note that for x 6= 0, L
6= 0, which is a contradiction. Hence, we can not
find a δ such that
for
every > 0, if 0 < |x − 0| < δ , then
1
Therefore, lim
does not exist.
x→0 x
1
x
− L < .
- Exercise 1.11 Prove that lim (3 + 2x) = 1
x→−1
- Exercise 1.12 Establish the following limits by means of Definition ??:
1. lim (5x + 7) = 17
x→2
lim (10 − 2x) = 16
x
3. lim
− 2 = − 35
x→1 3
4. lim x = π
x→π
5. lim 7 = 7
2.
x→−3
x→5
ASSESSMENT TEST
NAME:
SECTION:
SCORE:
Show that each of the following limits is true:
1. lim (2x + 1) = 1
t→0
2. lim (3 − 4x) = −1
x→1
t3 + 8
= 12
t→−2 t + 2
3. lim
SIGNATURE:
DATE:
Limit Theorems
- It would be a difficult task to solve each problem on limits using the
Definition 1.7. This section introduces theorems which may be used(as
shortcuts) to simplify the process of evaluating limits.
Limit Theorems
- It would be a difficult task to solve each problem on limits using the
Definition 1.7. This section introduces theorems which may be used(as
shortcuts) to simplify the process of evaluating limits.
- Theorem 1.13 (Uniqueness of a Limit)
If lim f (x) = L1 and lim f (x) = L2 , then L1 = L2 .
x→a
x→a
Limit Theorems
- It would be a difficult task to solve each problem on limits using the
Definition 1.7. This section introduces theorems which may be used(as
shortcuts) to simplify the process of evaluating limits.
- Theorem 1.13 (Uniqueness of a Limit)
If lim f (x) = L1 and lim f (x) = L2 , then L1 = L2 .
x→a
x→a
- Theorem 1.14 (Limit of a Constant Function)
If k is a constant, then for any real number a,
Limit Theorems
- It would be a difficult task to solve each problem on limits using the
Definition 1.7. This section introduces theorems which may be used(as
shortcuts) to simplify the process of evaluating limits.
- Theorem 1.13 (Uniqueness of a Limit)
If lim f (x) = L1 and lim f (x) = L2 , then L1 = L2 .
x→a
x→a
- Theorem 1.14 (Limit of a Constant Function)
If k is a constant, then for any real number a,
lim k = k.
x→a
Illustration: lim (−4) = −4.
x→2
Illustration: lim (−4) = −4.
x→2
(Discuss first limit of identity function)
- Corollary 1.15 If k is a constant and lim f (x) = L, then
x→a
Illustration: lim (−4) = −4.
x→2
(Discuss first limit of identity function)
- Corollary 1.15 If k is a constant and lim f (x) = L, then
x→a
lim kf (x) = k lim f (x) = kL.
x→a
x→a
Illustration: lim (−4) = −4.
x→2
(Discuss first limit of identity function)
- Corollary 1.15 If k is a constant and lim f (x) = L, then
x→a
lim kf (x) = k lim f (x) = kL.
x→a
x→a
Illustration: lim −2x = −2 lim x = −2(3) = −6.
x→3
x→3
Illustration: lim (−4) = −4.
x→2
(Discuss first limit of identity function)
- Corollary 1.15 If k is a constant and lim f (x) = L, then
x→a
lim kf (x) = k lim f (x) = kL.
x→a
x→a
Illustration: lim −2x = −2 lim x = −2(3) = −6.
x→3
x→3
- Theorem 1.16 (Limit of a Linear Function)
If m, b are constants, then
lim (mx + b) = ma + b.
x→a
- Proof: Consider the following cases:
- Proof: Consider the following cases:
Case 1: Suppose m = 0. Then mx + b = b and by Theorem 1.14,
lim b = b.
x→a
- Proof: Consider the following cases:
Case 1: Suppose m = 0. Then mx + b = b and by Theorem 1.14,
lim b = b.
x→a
Case 2: Suppose m 6= 0. If we let f (x) = mx + b and L = ma + b, then
by Definition 1.7, we must show that for every > 0 there exists a
number δ > 0 such that if 0 < |x − a| < δ , then
|(mx + b) − (ma + b)| < .
- Proof: Consider the following cases:
Case 1: Suppose m = 0. Then mx + b = b and by Theorem 1.14,
lim b = b.
x→a
Case 2: Suppose m 6= 0. If we let f (x) = mx + b and L = ma + b, then
by Definition 1.7, we must show that for every > 0 there exists a
number δ > 0 such that if 0 < |x − a| < δ , then
|(mx + b) − (ma + b)| < .
Take δ =
. Then
|m|
- Proof: Consider the following cases:
Case 1: Suppose m = 0. Then mx + b = b and by Theorem 1.14,
lim b = b.
x→a
Case 2: Suppose m 6= 0. If we let f (x) = mx + b and L = ma + b, then
by Definition 1.7, we must show that for every > 0 there exists a
number δ > 0 such that if 0 < |x − a| < δ , then
|(mx + b) − (ma + b)| < .
Take δ =
. Then
|m|
|(mx + b) − (ma + b)| = |mx − ma| = |m(x − a)|
- Proof: Consider the following cases:
Case 1: Suppose m = 0. Then mx + b = b and by Theorem 1.14,
lim b = b.
x→a
Case 2: Suppose m 6= 0. If we let f (x) = mx + b and L = ma + b, then
by Definition 1.7, we must show that for every > 0 there exists a
number δ > 0 such that if 0 < |x − a| < δ , then
|(mx + b) − (ma + b)| < .
Take δ =
. Then
|m|
|(mx + b) − (ma + b)| = |mx − ma| = |m(x − a)|
= |m||(x − a)| < |m|δ = |m| ·
= .
|m|
This implies that given > 0, if 0 < |x − a| < δ , where δ =
|(mx + b) − (ma + b)| < .
, then
|m|
This implies that given > 0, if 0 < |x − a| < δ , where δ =
|(mx + b) − (ma + b)| < .
Hence,
lim (mx + b) = ma + b.
x→a
, then
|m|
This implies that given > 0, if 0 < |x − a| < δ , where δ =
|(mx + b) − (ma + b)| < .
Hence,
lim (mx + b) = ma + b.
x→a
Illustration: lim (2y − 8) = 2(−1) − 8 = −2 − 8 = −10.
y→−1
, then
|m|
This implies that given > 0, if 0 < |x − a| < δ , where δ =
|(mx + b) − (ma + b)| < .
Hence,
lim (mx + b) = ma + b.
x→a
Illustration: lim (2y − 8) = 2(−1) − 8 = −2 − 8 = −10.
y→−1
- Corollry 1.17 (Limit of an Identity Function)
For any real number a,
, then
|m|
This implies that given > 0, if 0 < |x − a| < δ , where δ =
|(mx + b) − (ma + b)| < .
Hence,
lim (mx + b) = ma + b.
x→a
Illustration: lim (2y − 8) = 2(−1) − 8 = −2 − 8 = −10.
y→−1
- Corollry 1.17 (Limit of an Identity Function)
For any real number a,
lim x = a.
x→a
, then
|m|
This implies that given > 0, if 0 < |x − a| < δ , where δ =
|(mx + b) − (ma + b)| < .
Hence,
lim (mx + b) = ma + b.
x→a
Illustration: lim (2y − 8) = 2(−1) − 8 = −2 − 8 = −10.
y→−1
- Corollry 1.17 (Limit of an Identity Function)
For any real number a,
lim x = a.
x→a
Illustration: lim x = −2.
x→−2
, then
|m|
- Theorem 1.18 (Limit of a Sum)
If lim f (x) = L and lim g(x) = M , then
x→a
x→a
- Theorem 1.18 (Limit of a Sum)
If lim f (x) = L and lim g(x) = M , then
x→a
x→a
lim [f (x) + g(x)] = L + M.
x→a
That is,
- Theorem 1.18 (Limit of a Sum)
If lim f (x) = L and lim g(x) = M , then
x→a
x→a
lim [f (x) + g(x)] = L + M.
x→a
That is,
lim [f (x) + g(x)] = lim f (x) + lim g(x)
x→a
x→a
(Limit of a sum is the sum of the limits.)
x→a
- Theorem 1.18 (Limit of a Sum)
If lim f (x) = L and lim g(x) = M , then
x→a
x→a
lim [f (x) + g(x)] = L + M.
x→a
That is,
lim [f (x) + g(x)] = lim f (x) + lim g(x)
x→a
x→a
x→a
(Limit of a sum is the sum of the limits.)
Illustration:
lim [(2x − 7) + (4 − 3x)] = lim (2x − 7) + lim (4 − 3x)
x→1
x→1
x→1
- Theorem 1.18 (Limit of a Sum)
If lim f (x) = L and lim g(x) = M , then
x→a
x→a
lim [f (x) + g(x)] = L + M.
x→a
That is,
lim [f (x) + g(x)] = lim f (x) + lim g(x)
x→a
x→a
x→a
(Limit of a sum is the sum of the limits.)
Illustration:
lim [(2x − 7) + (4 − 3x)] = lim (2x − 7) + lim (4 − 3x)
x→1
x→1
x→1
= [2(1) − 7] + [4 − 3(1)] = −5 + 1 = −4.
- Theorem 1.19 (Limit of a Product)
- Theorem 1.19 (Limit of a Product)
If lim f (x) = L and lim g(x) = M , then
x→a
x→a
- Theorem 1.19 (Limit of a Product)
If lim f (x) = L and lim g(x) = M , then
x→a
x→a
lim [f (x) · g(x)] = L · M.
x→a
- Theorem 1.19 (Limit of a Product)
If lim f (x) = L and lim g(x) = M , then
x→a
x→a
lim [f (x) · g(x)] = L · M.
x→a
That is,
- Theorem 1.19 (Limit of a Product)
If lim f (x) = L and lim g(x) = M , then
x→a
x→a
lim [f (x) · g(x)] = L · M.
x→a
That is,
lim [f (x) · g(x)] = lim f (x) · lim g(x).
x→a
x→a
x→a
(Limit of a product is the product of the limits.)
- Theorem 1.19 (Limit of a Product)
If lim f (x) = L and lim g(x) = M , then
x→a
x→a
lim [f (x) · g(x)] = L · M.
x→a
That is,
lim [f (x) · g(x)] = lim f (x) · lim g(x).
x→a
x→a
x→a
(Limit of a product is the product of the limits.)
Illustration:
lim [(2x − 7)(−5x + 1)] = lim (2x − 7) lim (−5x + 1)
x→0
x→0
x→0
- Theorem 1.19 (Limit of a Product)
If lim f (x) = L and lim g(x) = M , then
x→a
x→a
lim [f (x) · g(x)] = L · M.
x→a
That is,
lim [f (x) · g(x)] = lim f (x) · lim g(x).
x→a
x→a
x→a
(Limit of a product is the product of the limits.)
Illustration:
lim [(2x − 7)(−5x + 1)] = lim (2x − 7) lim (−5x + 1)
x→0
x→0
x→0
= [2(0) − 7][−5(0) + 1] = (−7)(1) = −7.
- Theorem 1.20 (Limit of a Quotient)
- Theorem 1.20 (Limit of a Quotient)
If lim f (x) = L and lim g(x) = M , M 6= 0, then
x→a
x→a
- Theorem 1.20 (Limit of a Quotient)
If lim f (x) = L and lim g(x) = M , M 6= 0, then
x→a
x→a
L
f (x)
=
.
x→a g(x)
M
lim
- Theorem 1.20 (Limit of a Quotient)
If lim f (x) = L and lim g(x) = M , M 6= 0, then
x→a
x→a
L
f (x)
=
.
x→a g(x)
M
lim
That is,
- Theorem 1.20 (Limit of a Quotient)
If lim f (x) = L and lim g(x) = M , M 6= 0, then
x→a
x→a
L
f (x)
=
.
x→a g(x)
M
lim
That is,
lim f (x)
f (x)
= x→a
x→a g(x)
lim g(x)
lim
x→a
- Theorem 1.20 (Limit of a Quotient)
If lim f (x) = L and lim g(x) = M , M 6= 0, then
x→a
x→a
L
f (x)
=
.
x→a g(x)
M
lim
That is,
lim f (x)
f (x)
= x→a
x→a g(x)
lim g(x)
lim
x→a
(Limit of a quotient is the quotient of the limits.)
lim (x − 2)
3−2
x−2
= x→3
=
= 91 .
x→3 2x + 3
lim (2x + 3)
2(3) + 3
Illustration: lim
x→3
lim (x − 2)
3−2
x−2
= x→3
=
= 91 .
x→3 2x + 3
lim (2x + 3)
2(3) + 3
x→3
The following theorems give us the extended Theorem 1.18 and extended
Theorem 1.19 respectively:
Illustration: lim
lim (x − 2)
3−2
x−2
= x→3
=
= 91 .
x→3 2x + 3
lim (2x + 3)
2(3) + 3
x→3
The following theorems give us the extended Theorem 1.18 and extended
Theorem 1.19 respectively:
- Theorem 1.21 If lim f1 (x) = L1 , lim f2 (x) = L2 ,... and lim fn (x) = Ln ,
x→a
x→a
x→a
then
Illustration: lim
lim (x − 2)
3−2
x−2
= x→3
=
= 91 .
x→3 2x + 3
lim (2x + 3)
2(3) + 3
x→3
The following theorems give us the extended Theorem 1.18 and extended
Theorem 1.19 respectively:
- Theorem 1.21 If lim f1 (x) = L1 , lim f2 (x) = L2 ,... and lim fn (x) = Ln ,
x→a
x→a
x→a
then
Illustration: lim
lim [f1 (x) + f2 (x) + ... + fn (x)] = L1 + L2 + ... + Ln .
x→a
lim (x − 2)
3−2
x−2
= x→3
=
= 91 .
x→3 2x + 3
lim (2x + 3)
2(3) + 3
x→3
The following theorems give us the extended Theorem 1.18 and extended
Theorem 1.19 respectively:
- Theorem 1.21 If lim f1 (x) = L1 , lim f2 (x) = L2 ,... and lim fn (x) = Ln ,
x→a
x→a
x→a
then
Illustration: lim
lim [f1 (x) + f2 (x) + ... + fn (x)] = L1 + L2 + ... + Ln .
x→a
- Theorem 1.22 If lim f1 (x) = L1 , lim f2 (x) = L2 ,... and lim fn (x) = Ln ,
x→a
then
x→a
x→a
- Theorem 1.22 If lim f1 (x) = L1 , lim f2 (x) = L2 ,... and lim fn (x) = Ln ,
x→a
x→a
then
lim [f1 (x)f2 (x)...fn (x)] = L1 L2 ...Ln .
x→a
x→a
Other Limit Theorems
- Theorem 1.23 If lim f (x) = L and n is any positive integer, then we have
x→a
Other Limit Theorems
- Theorem 1.23 If lim f (x) = L and n is any positive integer, then we have
x→a
lim [f (x)]n = [lim f (x)]n = Ln .
x→a
x→a
Other Limit Theorems
- Theorem 1.23 If lim f (x) = L and n is any positive integer, then we have
x→a
lim [f (x)]n = [lim f (x)]n = Ln .
x→a
x→a
Illustration:
h
i5
lim (2x−12)5 = lim (2x − 12) = [2(5)−12]5 = [10−12]5 = (−2)5 = −32.
x→5
x→5
Other Limit Theorems
- Theorem 1.23 If lim f (x) = L and n is any positive integer, then we have
x→a
lim [f (x)]n = [lim f (x)]n = Ln .
x→a
x→a
Illustration:
h
i5
lim (2x−12)5 = lim (2x − 12) = [2(5)−12]5 = [10−12]5 = (−2)5 = −32.
x→5
x→5
- Corollary 1.24 lim xn = an .
x→a
Other Limit Theorems
- Theorem 1.23 If lim f (x) = L and n is any positive integer, then we have
x→a
lim [f (x)]n = [lim f (x)]n = Ln .
x→a
x→a
Illustration:
h
i5
lim (2x−12)5 = lim (2x − 12) = [2(5)−12]5 = [10−12]5 = (−2)5 = −32.
x→5
x→5
- Corollary 1.24 lim xn = an .
x→a
Proof: This follows from Theorem 1.23, if we let f (x) = x.
- Theorem 1.25 If f is a polynomial function, then
lim f (x) = f (a)
x→a
for every real number a.
- Theorem 1.25 If f is a polynomial function, then
lim f (x) = f (a)
x→a
for every real number a.
Proof: Let f (x) = bn xn + bn−1 xn−1 + ... + b3 x3 + b3 x2 + b1 x + b0 , where
bi ’s are real numbers. Then applying Theorem 1.18, Theorem 1.23 and
Corollary 1.15,
- Theorem 1.25 If f is a polynomial function, then
lim f (x) = f (a)
x→a
for every real number a.
Proof: Let f (x) = bn xn + bn−1 xn−1 + ... + b3 x3 + b3 x2 + b1 x + b0 , where
bi ’s are real numbers. Then applying Theorem 1.18, Theorem 1.23 and
Corollary 1.15,
lim f (x) = lim bn xn + bn−1 xn−1 + ... + b3 x3 + b3 x2 + b1 x + b0
x→a
x→a
= bn an + bn−1 an−1 + ... + b3 a3 + b3 a2 + b1 a + b0 = f (a).
- Theorem 1.26 If lim f (x) = L, then
x→a
- Theorem 1.26 If lim f (x) = L, then
x→a
lim
x→a
p
n
f (x) =
q
n
lim f (x) =
x→a
√
n
L,
- Theorem 1.26 If lim f (x) = L, then
x→a
lim
x→a
provided
√
n
L ∈ R.
p
n
f (x) =
q
n
lim f (x) =
x→a
√
n
L,
- Theorem 1.26 If lim f (x) = L, then
x→a
lim
x→a
p
n
f (x) =
q
n
lim f (x) =
x→a
√
n
L,
√
n
provided L ∈ R.
Illustration:
q
p
√
√
lim 3 −8 + 21x = 3 lim (−8 + 21x) = 3 −8 + 21(0) = 3 −8 = −2.
x→0
x→0
- Theorem 1.26 If lim f (x) = L, then
x→a
lim
x→a
p
n
f (x) =
q
n
lim f (x) =
x→a
√
n
L,
√
n
provided L ∈ R.
Illustration:
q
p
√
√
lim 3 −8 + 21x = 3 lim (−8 + 21x) = 3 −8 + 21(0) = 3 −8 = −2.
x→0
x→0
- Corollary 1.27 If q is a rational function and a is in the domain of q , then
- Theorem 1.26 If lim f (x) = L, then
x→a
lim
x→a
p
n
f (x) =
q
n
lim f (x) =
x→a
√
n
L,
√
n
provided L ∈ R.
Illustration:
q
p
√
√
lim 3 −8 + 21x = 3 lim (−8 + 21x) = 3 −8 + 21(0) = 3 −8 = −2.
x→0
x→0
- Corollary 1.27 If q is a rational function and a is in the domain of q , then
lim q(x) = q(a).
x→a
- Theorem 1.28 If F and G are two functions such that F (x) = G(x) for
all x 6= a, and if lim G(x) exists, then lim F (x) exists. Moreover,
x→a
x→a
- Theorem 1.28 If F and G are two functions such that F (x) = G(x) for
all x 6= a, and if lim G(x) exists, then lim F (x) exists. Moreover,
x→a
x→a
lim F (x) = lim G(x).
x→a
x→a
- Theorem 1.28 If F and G are two functions such that F (x) = G(x) for
all x 6= a, and if lim G(x) exists, then lim F (x) exists. Moreover,
x→a
x→a
lim F (x) = lim G(x).
x→a
x2 −x
x→1 x−1
Illustration: lim
= lim
x→1
x(x−1)
x−1
x→a
= lim x = 1,
x→1
- Theorem 1.28 If F and G are two functions such that F (x) = G(x) for
all x 6= a, and if lim G(x) exists, then lim F (x) exists. Moreover,
x→a
x→a
lim F (x) = lim G(x).
x→a
x2 −x
x→1 x−1
Illustration: lim
where
x2 −x
x−1
= lim
x→1
x(x−1)
x−1
= x for all x 6= 1.
x→a
= lim x = 1,
x→1
- Note that all the illustrations that we have can be done by direct
substitution. So, the first thing to do to evaluate limit of a function is
direct substitution. If the result is a real number, then that real number
is the limit of the function. However, if the result is 00 , we may simplify
the expression by factoring or rationalizing, whichever is applicable.
- Note that all the illustrations that we have can be done by direct
substitution. So, the first thing to do to evaluate limit of a function is
direct substitution. If the result is a real number, then that real number
is the limit of the function. However, if the result is 00 , we may simplify
the expression by factoring or rationalizing, whichever is applicable.
x2 + 2x
- Illustration: lim
x→−2 x + 2
- Note that all the illustrations that we have can be done by direct
substitution. So, the first thing to do to evaluate limit of a function is
direct substitution. If the result is a real number, then that real number
is the limit of the function. However, if the result is 00 , we may simplify
the expression by factoring or rationalizing, whichever is applicable.
x2 + 2x
- Illustration: lim
x→−2 x + 2
By direct substitution,
x2 + 2x
(−2)2 + 2(−2)
0
=
= .
x→−2 x + 2
−2 + 2
0
lim
- Note that all the illustrations that we have can be done by direct
substitution. So, the first thing to do to evaluate limit of a function is
direct substitution. If the result is a real number, then that real number
is the limit of the function. However, if the result is 00 , we may simplify
the expression by factoring or rationalizing, whichever is applicable.
x2 + 2x
- Illustration: lim
x→−2 x + 2
By direct substitution,
x2 + 2x
(−2)2 + 2(−2)
0
=
= .
x→−2 x + 2
−2 + 2
0
lim
Thus,
x2 + 2x
x(x + 2)
= lim
= lim x = −2.
x→−2 x + 2
x→−2 x + 2
x→−2
lim
√
x−3
9−3
3−3
0
=
=
= .
Illustration: lim
x→9 9 − x
9−9
9−9
0
√
√
x−3
9−3
3−3
0
=
=
= .
Illustration: lim
x→9 9 − x
9−9
9−9
0
Now, rationalizing the numerator yields
√
√
√
x−3
x−3
x+3
= lim
·√
lim
x→9 9 − x
x→9 9 − x
x+3
√
√
x−3
9−3
3−3
0
=
=
= .
Illustration: lim
x→9 9 − x
9−9
9−9
0
Now, rationalizing the numerator yields
√
√
√
x−3
x−3
x+3
= lim
·√
lim
x→9 9 − x
x→9 9 − x
x+3
x−9
−1
−1
√
= lim
= lim √
=
.
x→9 (9 − x)( x + 3)
x→9
6
x+3
√