BASIC

CALCULUS
 Topic 1:

Introduction to limits
Learning objectives:

1. Define Limit of a function: lim 𝑓 𝑥 = 𝐿

𝑥→𝑐
2. Compare limit of a function and the value of a
function at C
lim 𝑓 𝑥 = 𝐿 versus f (c)
𝑥→𝑐
Definition
 CALCULUS

• The word calculus is a latin word, meaning

originally “small pebble”.
• Isaac Newton and Gottfried Wilhelm Leibniz (17th
century)
• Mathematical study of continuous change.
Limit of a
Function
 LIMIT

• A point or level beyond which something does not

or may not extend or pass.
• Limit is the value that a function approaches as the
input approaches some value.
Read as: ‘‘The LIMIT
limit of f(x) as
x approaches
c is L.”
𝐥𝐢𝐦 𝒇 𝒙 = 𝑳
𝒙→𝒄 Limit denoted by L is
the unique real value
constant that may
that f(x) will approach
or may not be in
as x approaches c.
the domain of
function
Examples:
Let’s find the Limit of a Function of the
following given:
2
1. lim 1 + 3𝑥 4. lim
𝑥 −7𝑥+6
𝑥→2 𝑥→6 𝑥−6

2. lim 𝑥 2 −3
𝑥→0 5. lim 𝑓(𝑥)
𝑥→4
3. lim 𝑥
{ }
𝑓 𝑥 = 𝑥 + 1 𝑖𝑓 𝑥 < 4
𝑥→0 2
𝑥 − 4 + 3 𝑖𝑓 𝑥 ≥ 4
 1. lim 1 + 3𝑥
𝑥→2
Approaching 2 from its right Approaching 2 from its left

𝑥 𝑓(𝑥) 𝑥 𝑓(𝑥)
1 4 3 10
1.4 5.2 2.5 8.5
1.7 6.1 2.2 7.6
1.9 6.7 2.1 7.3
1.95 6.85 2.03 7.09
1.997 6.991 2.009 7.027
1.9999 6.9997
lim 1 + 3𝑥 = 7 2.0005 7.0015
𝑥→2
1.99999996.9999997 2.0000001 7.0000003
lim− 1 + 3𝑥 = 7
𝑥→2
lim 1 + 3𝑥 = 7 lim+ 1 + 3𝑥 = 7
𝑥→2
𝑥→2
 2
2. lim 𝑥 − 3
𝑥→0
Approaching 0 from its left Approaching 0 from its right

𝑥 𝑓(𝑥) 𝑥 𝑓(𝑥)
−0.5 −2.75 0.5 −2.75
−0.2 −2.96 0.2 −2.96
−0.005 −2.999975 0.005 −2.999975
−0.00009 −2.999999992 0.00009 −2.999999992

lim 𝑥 2 − 3 = −3
𝑥→0
 2
lim− 𝑥 − 3 = −3 lim 𝑥 2 − 3 = −3 lim+ 𝑥 2 − 3 = −3
𝑥→0 𝑥→0
𝑥→0
 3. lim 𝑥
𝑥→0
Approaching 0 from its left Approaching 0 from its right

𝑥 𝑓(𝑥) 𝑥 𝑓(𝑥)
−0.3 0.3 0.3 0.3
−0.01 0.01 0.01 0.01
−0.00009 0.00009 0.00009 0.00009
−0.00000001 0.00000001 0.00000001 0.00000001

lim 𝑥 = 0
𝑥→0
lim− 𝑥 = 0
𝑥→0
lim 𝑥 = 0 lim+ 𝑥 = 0
𝑥→0 𝑥→0
 2
𝑥 − 7𝑥 + 6
4. lim
𝑥→6 𝑥−6
Approaching 0 from its left Approaching 0 from its right

𝑥 𝑓(𝑥) 𝑥 𝑓(𝑥)
5 4 7 6
5.025 4.025 6.025 5.025
5.995 4.995 6.0009 5.0009
5.9999 4.9999 6.00001 5.00001

2
𝑥 − 7𝑥 + 6
lim =5
𝑥→6 𝑥−6
 2
𝟐
𝒙 − 𝟕𝒙 + 𝟔 𝑥 − 7𝑥 + 6 𝒙𝟐 − 𝟕𝒙 + 𝟔
lim− =5 lim =5 lim+
𝒙−𝟔
=5
𝑥→6 𝒙−𝟔 𝑥→6 𝑥−6 𝑥→6
 5. lim 𝑓(𝑥) {
𝑓 𝑥 = 𝑥 + 1 𝑖𝑓 𝑥 < 4
2
}
𝑥→4 𝑥 − 4 + 3 𝑖𝑓 𝑥 ≥ 4

Approaching 4 from its left Approaching 4 from its right

𝟐
𝒇 𝒙 =𝒙+𝟏 𝒇 𝒙 = 𝒙 − 𝟒 +3
𝑥 𝑓(𝑥) 𝑥 𝑓(𝑥)
3.7 4.7 4.3 3.09
3.85 4.85 4.1 3.01
3.995 4.995 4.001 3.000001
3.99999 4.99999 4.00001 3.0000000001

lim 𝑓(𝑥) 𝐷𝑂𝐸𝑆 𝑁𝑂𝑇 𝐸𝑋𝐼𝑆𝑇 (𝐷𝑁𝐸)

𝑥→4
lim 𝑓(𝑥) 𝐷𝑂𝐸𝑆 𝑁𝑂𝑇 𝐸𝑋𝐼𝑆𝑇 (𝐷𝑁𝐸)
𝑥→4
lim− 𝑓 𝑥 = 5
𝑥→4
≠ lim+ 𝑓(𝑥) = 3
𝑥→4
 REMEMBER:
• If the limit of f(x) does not exist, DO NOT use the word equals nor write
the symbol 𝒍𝒊𝒎 𝒇 𝒙 = 𝑫𝑵𝑬
𝒙→𝒄

• This direction may be specified in the limit notation, lim 𝑓 𝑥 = 𝐿 by

𝑥→𝑐
adding certain symbols.
LEFT : 𝒍𝒊𝒎− 𝒇 𝒙 = 𝑳 RIGHT: 𝒍𝒊𝒎+ 𝒇 𝒙 = 𝑳
𝒙→𝒄 𝒙→𝒄
 REMEMBER:
• lim 𝑓 𝑥 𝑒𝑥𝑖𝑠𝑡 𝒊𝒇 𝒂𝒏𝒅 𝒐𝒏𝒍𝒚 𝒊𝒇
𝑥→𝑐
𝒍𝒊𝒎− 𝒇 𝒙 = 𝒍𝒊𝒎+ 𝒇 𝒙
𝒙→𝒄 𝒙→𝒄
• lim 𝑓 𝑥 𝑫𝑵𝑬,
𝑥→𝑐
𝒍𝒊𝒎− 𝒇 𝒙 ≠ 𝒍𝒊𝒎+ 𝒇 𝒙
𝒙→𝒄 𝒙→𝒄

• 𝒍𝒊𝒎− 𝒇 𝒙 = 𝑳 𝑎𝑛𝑑 𝒍𝒊𝒎+ 𝒇 𝒙 = 𝑳

𝒙→𝒄 𝒙→𝒄
are also referred to as one-sided limits
 Limit of a
Function at c
VS.
the Value of the
Function at c
Value of the Function at c

lim 1 + 3𝑥 Limit of a Function

𝑥→2
1. lim 1 + 3𝑥 = 7
𝑓 2 = 1 + 3𝑥 𝑥→2

𝑓 2 = 1 + 3(2)
𝑓 2 =7
Value of the Function at c
2 Limit of a Function
lim 𝑥 − 3
𝑥→0 2
2. lim 𝑥 − 3 = −3
2 𝑥→0
𝑓 0 = 𝑥 −3
2
𝑓 0 = (0) − 3
𝑓 0 = −3
Value of the Function at c
Limit of a Function
lim 𝑥 3. lim 𝑥 = 0
𝑥→0 𝑥→0

𝑓 0 = 𝑥
𝑓 0 = 0
𝑓 0 =0
Value of the Function at c
2
𝑥 − 7𝑥 + 6 Limit of a Function
lim
𝑥→6 𝑥−6 4. lim
𝑥 2 −7𝑥+6
=5
𝑥→6 𝑥−6

2
𝑥 − 7𝑥 + 6
𝑓 6 =
𝑥−6
𝑓 6 𝑖𝑠 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
Value of the Function at c
lim 𝑓 𝑥 Limit of a Function lim 𝑓 𝑥 𝐷𝑁𝐸
𝑥→4 𝑥→4
2 5. 𝑓 𝑥 =
𝑓 4 = 𝑥−4 +3
{ }
𝑥 + 1 𝑖𝑓 𝑥 < 4
2
𝑥 − 4 + 3 𝑖𝑓 𝑥 ≥ 4
2
𝑓 4 = 4−4 +3
𝑓 4 =0+3
𝑓 4 =3
 Limit
Theorems
1. Limit of a constant
- The limit of a constant is itself.

1. lim𝑥→𝑐 7 = 7
2. lim𝑥→𝑐 121 = 121
3. lim𝑥→𝑐 -25 = - 25
4. lim𝑥→𝑐 1 = 1
 2. Limit of x as x approaches c

- The limit of x as x approaches c is equal to c.

1. lim𝑥→9 𝑥 = 9
2. lim𝑥→−3 𝑥 = -3
3. lim𝑥→0 𝑥 = 0
4. lim𝑥→−2 𝑥 = - 2
 Constant Multiple
Theorem
- The limit of a multiple of a function is equal to the product of
constant and limit of the function.
➢ lim 𝑘 𝑓(𝑥) = 𝑘 ∗ lim 𝑓(𝑥) = 𝑘 ∗ 𝐿
𝑥→𝑐
= 3 ∗ lim 𝑓(𝑥)
If lim 𝑓 𝑥 = 25 then, 𝑥→𝑐
𝑥→𝑐
= 3(25)
1. lim 3 𝑓(𝑥)
𝑥→𝑐 lim 3 𝑓(𝑥) = 75
𝑥→𝑐
Constant Multiple Theorem

➢ lim 𝑘 𝑓(𝑥) = 𝑘 ∗ lim 𝑓(𝑥) = 𝑘 ∗ 𝐿

𝑥→𝑐

lim𝑥→3 8x
1. lim 8x =8∗3
𝑥→3

= 8 ∗ lim x lim 8x = 24
𝑥→3 𝑥→3
Addition/Subtraction
Theorem
- The limits of the sum of the function is equal to the sum of the
limit of each function.

lim f (x) + g(x) = lim f (x) + lim g(x)

𝑥→𝑐 𝑥→𝑐 𝑥→𝑐

lim f (x) − g(x) = lim f (x) − lim g(x)

𝑥→𝑐 𝑥→𝑐 𝑥→𝑐
Addition/Subtraction
Theorem
If 𝑙𝑖𝑚 f (x) = 2 and 𝑙𝑖𝑚 g(x) = 3
𝑥→𝑐 𝑥→𝑐

1. lim f (x) + g(x) 2. lim f (x) − g(x)

𝑥→𝑐 𝑥→𝑐

= lim f (x) + lim g(x) = lim f (x) − lim g(x)

𝑥→𝑐 𝑥→𝑐 𝑥→𝑐 𝑥→𝑐
= 2 + 3 = 2 − 3
= 5 = −1
 Addition/Subtraction
Theorem
1. 𝑙𝑖𝑚 3𝑥 + 9
𝑥→1

= lim 3𝑥 + lim 9 =3∗1+9

𝑥→1 𝑥→1
=3+9
= 3 ∗ lim 𝑥 + lim 9
𝑥→1 𝑥→1
= 12
Multiplication Theorem

- The limits of the product of the functions is equal to

the product of the limit of each function
lim f (x) ∗ g(x) = lim f (x) ∗ lim g(x)
𝑥→𝑐 𝑥→𝑐 𝑥→𝑐

If lim f (x) = 4 and lim g(x) = 5

𝑥→𝑐 𝑥→𝑐
= lim f (x) ∗ g(x)
𝑥→𝑐
= lim f (x) ∗ lim g(x)
𝑥→𝑐 𝑥→𝑐
=4∗5
= 20
Division Theorem
- The limits of the quotient of the functions is equal to the
quotient of the limit of each function.

𝑓(𝑥) lim 𝑓(𝑥)

𝑥→𝑐
lim = , if lim 𝑔(𝑥) ≠ 0
𝑥→𝑐 𝑔(𝑥) lim 𝑔(𝑥) 𝑥→𝑐
𝑥→𝑐
 𝑓(𝑥) lim 𝑓(𝑥)
𝑥→𝑐
lim = , if lim 𝑔(𝑥) ≠ 0
𝑥→𝑐 𝑔(𝑥) lim 𝑔(𝑥) 𝑥→𝑐
𝑥→𝑐

If lim 𝑓(𝑥) = 10 and lim 𝑔 𝑥 = 2

𝑥→𝑐 𝑥→𝑐

𝑓(𝑥)
1. lim 10
𝑥→𝑐 𝑔(𝑥) =
2
lim 𝑓(𝑥)
𝑥→𝑐 =5
=
lim 𝑔(𝑥)
𝑥→𝑐
Power Theorem

- The limits of the function to the power of n is equal to

the limit of the function raised to the power of n.

𝑛 𝑛 𝑛
lim 𝑓 𝑥 = [lim 𝑓(𝑥)] =𝐿
𝑥→𝑐 𝑥→𝑐
 Power Theorem

𝑛 𝑛 𝑛
lim 𝑓 𝑥 = [lim 𝑓(𝑥)] =𝐿
𝑥→𝑐 𝑥→𝑐
3
1. lim 5𝑥 3
𝑥→2 = 5 ∗ [lim 𝑥]
𝑥→2
3
= 5 ∗ [ 2]
=5∗8
=40
 Power Theorem

𝑛 𝑛 𝑛
lim 𝑓 𝑥 = [lim 𝑓(𝑥)] =𝐿
𝑥→𝑐 𝑥→𝑐
−3
1. lim 2𝑥 −3 = 2 ∗ lim 𝑥
𝑥→1
𝑥→1
= 2 ∗ [lim 𝑥] −3 =2∗1
𝑥→1
−3 =2
= 2 ∗ [1]
 - The limits of the nth root of the
Radical/Root Theorem function is equal to the nth root of the
limit of the function.

𝑛 𝑛 𝑛
lim 𝑓(𝑥). = lim 𝑓(𝑥) = 𝐿
𝑥→𝑐 𝑥→𝑐

if lim 𝑓(𝑥) = 81
𝑥→𝑐 2
1. lim 𝑓(𝑥)
𝑥→𝑐
2
= lim 𝑓(𝑥)
𝑥→𝑐
2
= 81
=9
 Evaluating
Limits of
Polynomials
(EXAMPLES)
Using Limit Theorems
𝟑 𝟐
.𝐥𝐢𝐦(𝟑𝒙 + 𝟑) .𝐥𝐢𝐦. (𝒙 + 𝟐𝒙 - 4x – 3)
𝒙→𝟐 𝒙→𝟐 𝑥 2 −7𝑥+6
1. lim 1 + 3𝑥 = 7 4.
3 lim 2 = 5
𝑥→23𝑥 + lim (3)
= lim = lim . 𝑥 +𝑥→6
lim . 2𝑥
𝑥−6- lim . 4x – lim .3
𝑥→2 𝑥→2 𝑥→2 𝑥→2 𝑥→2 𝑥→2
2. lim 𝑥
− 2
3 = −3 2 .
= 3 𝑥→0
* lim 𝑥 + 3 = (2)3 + 2 *
5. limlim . -𝑥4 * lim . 𝑥- 3
𝑥→2 𝑥→2 𝑓 𝑥 𝐷𝑁𝐸 𝑥→2
𝑥→4
= 3(2)
3. lim+𝑥3 = 0 = 8 + 2(2)2 – 4(2) – 3

{ }
𝑥→0 𝑓 𝑥 = 𝑥 + 1 𝑖𝑓 𝑥 < 4
=6+3 =8+8–8–3
2
𝑥 − 4 + 3 𝑖𝑓 𝑥 ≥ 4
=9 =5
Using Limit Theorems
𝒙
𝐥𝐢𝐦
𝟐 𝒙→𝟑 (𝒙2 − 𝟐)
1. lim
𝒍𝒊𝒎1 + 3𝑥 = 7 4. lim
𝑥 −7𝑥+6
=5
𝑥→2
𝒙→𝟓 𝒙 lim 𝑥.
𝑥→ 3 𝑥→6 𝑥−6
.
2. lim 2
lim 𝑥
𝑥→5
. 2 − 3 = −3 = lim ( 𝑥 −2)
= 𝑥→0
. 𝑥→3 5. lim 𝑓 𝑥 𝐷𝑁𝐸
𝑥→4
lim 𝑥 3 3
𝑥→5
3. lim 𝑥 = 0 = =
lim 𝑥 −lim 2 3 − 2
{𝑥−4 }
2𝑥→0 𝑓 𝑥 = 𝑥 + 1 𝑖𝑓 𝑥 < 4
= 𝑥→3 𝑥→3
5 2 +3 3 𝑖𝑓 𝑥 ≥ 4
= =3
1
Using Limit Theorems
𝟑
1. 1..𝒍𝒊𝒎
lim 1 +𝒙3𝑥 = 7 2. 𝒍𝒊𝒎 𝒙 𝑥−2 −7𝑥+6
𝟐𝟕
𝒙→𝟒
𝑥→2 4. lim
𝒙→𝟎 =5
𝑥−6
𝑥→6
2 3
2. =limlim
𝑥 −𝑥 3 = −3 = lim 𝑥 − lim 27
𝑥→0𝑥→4 5.𝑥→0lim 𝑓𝑥→0
𝑥 𝐷𝑁𝐸
𝑥→4
3
3. =lim4 𝑥 = 0 = 0 − 27
{ }
𝑥→0 𝑓 𝑥 = 𝑥 + 1 𝑖𝑓 𝑥 < 4
=2 = - 3 𝑥 − 4 2 + 3 𝑖𝑓 𝑥 ≥ 4
Using Substitution
1. 𝐥𝐢𝐦 𝟓𝐱𝟒 − 𝟒𝐱𝟑 + 𝟑𝒙𝟐 − 𝟐𝒙 + 𝟑
𝒙→𝟏

Let: x = 1

= lim 5 1 4 −4 1 3 +3 1 2 −2 1 +3
𝑥→1

=5–4+3–2+3

=5
Using Substitution
2. 𝐥𝐢𝐦 𝟒𝐭𝟑 − 𝟓𝐭𝟐 + 𝒕 + 𝟔
𝒕→𝟐

Let: t = 2

= lim 4 2 3 −5 2 2 +2+6
𝑡→2

= 32 – 20 + 2 + 6

= 20
Thank
you