Module 2
LIMITS OF FUNCTIONS
Differential and integral Calculus
Lesson 1:
LIMITS OF FUNCTIONS
Differential and integral Calculus
OBJECTIVES:
• define limits
• evaluate limits
Differential and integral Calculus
DEFINITION: Limits
The most basic use of limits is to describe how a
function behaves as the independent variable
approaches a given value. For example let us
2
examine the behavior of the function f ( x ) = x − x + 1
for x-values closer and closer to 2. It is evident from
the graph and the table in the next slide that the
values of f(x) get closer and closer to 3 as the values
of x are selected closer and closer to 2 on either the
left or right side of 2. We describe this by saying
that the “limit of f ( x ) = x 2 − x + 1 is 3 as x
approaches 2 from either side,” we write
(
)
lim x 2 −Differential
x + 1and=integral
3 Calculus
x →2
y
y = x2 − x + 1
f(x)
3
f(x)
O
x
2
x
1.9
1.95
1.99
1.995 1.999
F(x)
2.71
2.852
2.97
2.985 2.997
2
2.001 2.005
2.05
2.1
3.003 3.015 3.031 3.152
3.31
left side
2.01
right side
Differential and integral Calculus
This leads us to the following general idea.
Differential and integral Calculus
EXAMPLE
Use numerical evidence to make a conjecture about
x −1
the value of lim
.
x →1
x −1
x −1
Although the function f ( x ) =
is undefined at
x −1
x=1, this has no bearing on the limit.
The table shows sample x-values approaching 1 from
the left side and from the right side. In both cases the
corresponding values of f(x) appear to get closer and
closer to 2, and hence we conjecture that
x −1
lim
= 2 and is consistent with the graph of f.
x →1
x −1
Differential and integral Calculus
x
.99
.999
.9999
.99999
F(x) 1.9949 1.9995 1.99995 1.999995
1
1.00001
1.0001
2.000005
2.00005 2.0005
Differential and integral Calculus
1.001
1.01
2.004915
THEOREMS ON LIMITS
Our strategy for finding limits algebraically has two
parts:
• First we will obtain the limits of some simpler
function
• Then we will develop a list of theorems that will
enable us to use the limits of simple functions as
building blocks for finding limits of more
complicated functions.
Differential and integral Calculus
We start with the following basic theorems,
which are illustrated in Fig 1.2.1
1.2.1 Theorem
Let a and k be real numbers.
(a) lim k = k
(b)lim x = a
x →a
x →a
Differential and integral Calculus
Figure 1.2.1
Differential and integral Calculus
Example 1.
If f (x ) = k is a constant function, then the values of f(x)
remain fixed at k as x varies, which explains why
f(x) → k as x → a for all values of a.
For example,
lim 3 = 3
lim 3 = 3
lim 3 = 3
x →-25
x →
x →0
Example 2.
𝐺𝑖𝑣𝑒𝑛 f 𝑥 = x, if x 𝑎𝑝𝑝𝑟oaches a then it must also be true that f 𝑥
𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ𝑒𝑠 a .
For example,
lim 𝑥 = 0
𝑥→0
lim 𝑥 = −2
𝑥→−2
lim 𝑥 = 𝜋
𝑥→𝜋
Differential and integral Calculus
The following theorem will be our basic tool for finding limits
algebraically.
Differential and integral Calculus
This theorem can be stated informally as follows:
a)
b)
c)
d)
The limit of a sum is the sum of the limits.
The limit of a difference is the difference of the limits.
The limits of a product is the product of the limits.
The limits of a quotient is the quotient of the limits,
provided the limit of the denominator is not zero.
e) The limit of the nth root is the nth root of the limit
provided the limit value is non-negative if n is even.
• A constant factor can be moved through a limit symbol.
Differential and integral Calculus
EXAMPLE : Evaluate the following limits.
1. lim (2 x + 5 ) = lim 2 x + lim 5
x →4
x →4
x →4
2. lim(6 x − 12) = lim 6 x − lim(12)
x →3
x →3
= 2 lim x + lim 5
x →4
x →4
= 2( 4 ) + 5
=8+5
= 13
3. lim (4 − x )( 5 x − 2 ) = lim (4 − x )  lim (5 x − 2 )
x →3
= 6( 3 ) − 12
= 18 - 12
=6
)(
)
(
= (lim 4 − lim x )  (5 lim x − lim 2 )
x →3
x →3
= lim 4 − lim x  lim 5 x − lim 2
x →3
x →3
x →3
x →3
x →3
x →3
= (4 − 3)(5( 3 ) − 2 )
= (1)(13)
= 13
Differential and integral Calculus
x →3
x →3
x →3
lim 2 x
2 lim x
2x
x →5
x →5
4. lim
=
=
x →5 5 x − 4
lim(5 x ) − lim(4 ) 5 lim(x ) − lim(4 )
x →5
x →5
x →5
2(5 )
=
25 − 4
10
=
21
(
8x + 1
8x + 1
6. lim
= lim
x →1
x →1 x + 3
x+3
x →3
9 3
=
=
4 2
)
5. lim(3 x + 6 ) = lim(3 x + 6 )
3
x →5
3
)
= (3 lim x + lim 6 )
(
x →3
= lim 3 x + lim 6
x →3
3
x →3
3
x →3
x →3
= ((3  3) + 6 ) = (15)
= 3375
3
3
Differential and integral Calculus
OR
When evaluating the limit of a function at a
given value, simply replace the variable by
the indicated limit then solve for the value of
the function:
(
)
lim 3 x + 4 x − 1 = 3 ( 3 ) + 4 ( 3 ) − 1
x →3
2
2
= 27 + 12 − 1
= 38
Differential and integral Calculus
EXAMPLE: Evaluate the following limits.
x3 + 8
1. lim
x →−2 x + 2
Solution:
x 3 + 8 (− 2 ) + 8 − 8 + 8 0
lim
=
=
=
(indeterminate)
x → −2 x + 2
−2+2
0
0
3
Equivalent function:
= lim
x →−2
( x + 2) ( x 2 − 2x + 4 )
x+2
= lim x 2 − 2 x + 4
x →−2
(
)
= ( −2 ) − 2 ( −2 ) + 4
2
= 4+4+4
= 12
x3 + 8
 lim
= 12
x →−2 x + 2
Differential and integral Calculus
Note: In evaluating a limit of a quotient which
reduces to 0 , simplify the fraction. Just remove
0
the common factor in the numerator and
denominator which makes the quotient 0 .
0
To do this use factoring or rationalizing the
numerator or denominator, wherever the radical is.
Differential and integral Calculus
x+2 − 2
2. lim
x →0
x
Solution:
lim
x →0
x+2 − 2
=
x
2− 2 0
=
(indeterminate)
0
0
Rationalizing the numerator:
lim
x →0
x+2− 2 x+2+ 2

= lim
x
x + 2 + 2 x →0 x
 lim
x →0
 lim
x →0
1
x+2 + 2
=
1
2+ 2
=
1
2 2
(
=
x +2−2
x+2 + 2
2
4
x+2 − 2
2
=
x
4
Differential and integral Calculus
)
8x 3 − 27
3. lim3
4x 2 − 9
x→
2
Solution:
3
8 x 3 − 27
lim3
=
2
x→2
4x − 9
3
8   − 27
2
2
3
4  − 9
2
=
27 − 27 0 (indeterminate)
=
9−9
0
Factoring the numerator and the denominator of the radicand:
8 x − 27
lim3
= lim3
2
x→2
x→2
4x − 9
3
(
2
2
x
−
3
4
x
+ 6x + 9
(
)
( 2x + 3 )( 2 x − 3 )
)
4x 2 + 6x + 9
9+9+9
= lim3
=
x→ 2
2x + 3
3+3
27
9
3
3 2
=
=
=
=
6 and 2integral Calculus
2
Differential
2
x3 + 2x + 3
4. lim
x→2
x2 + 5
Solution:
x + 2x + 3
=
2
x +5
3
lim
x →2
( 2) + 2 ( 2) + 3
3
( 2) + 5
2
8+4+3
=
4+5
=
15
9
=
15
3
Differential and integral Calculus
EXERCISE:
(
Evaluate the following limits.
1. lim 4 x − 5 x + 2
x →3
2
)
3x − 1
6. lim 2
1
x→ 9 x − 1
3
2x + 1
2. lim 2
x → −1 x − 3 x + 4
3. lim
x → −1
x 2 + 3x + 4
x3 + 1
 4 y3 + 8 y 

4. lim
y →2
 y+4 
x3 − 8
5. lim
x→2 x − 2
1
3
2x 3 + 3x 2 − 2x − 3
7. lim
x →1
x2 − 1
8.
(
y − 1)( y 2 + 2 y − 3)
lim
y →1
y2 − 2y + 1
(
)
9. lim 2 x 4 − 9 x 3 + 19
x →5
w2 + 7 w + 7
10. lim 2
w→ −1 w − 4 w − 5
Differential and integral Calculus
−
1
2