Chapter 2

Limits and Continuity

2.1 Limit of a Function

3x,3xxf

2 9( )

3

xf x

x

32 4

6

5

7

3x,3xxf

6

3

2 9( )

3

xf x

x

2 4

5

7

3x,3xxf

6

3

2 9( )

3

xf x

x

2 4

5

7

3x,3xxf

6

3

2 9( )

3

xf x

x

3

lim 6x

f x

The limit as x approaches

3 of f(x) 6.

2 4

5

7

As x approaches 3, f(x)

approaches 6.

Limit of a Function (an intuitive definition)

Let f be a function defined on some open interval

containing a except possibly at a itself. Then, the

limit of f as x approaches a is L, written as

limx a

f x L

if f(x) gets closer and closer to one and only one

number L as x takes the values that are closer to a.

Limit of a Function (an intuitive definition)

To say that means that: limx a

f x L

The distance (or difference, to be more specific)

between f(x) and L can be made arbitrarily small

by requiring that x be sufficiently close to but

different from a.

Remark: The small distances (or differences)

stated above must be quantified to have a more

precise definition of limit.

Limit of a Function

Let and be positive real numbers,

i.e. 0 and 0.

Think and as small positive numbers.

“the distance between f(x) and L can be made

arbitrarily small”:

f x L f x L

L f x L

Limit of a Function

Let and be positive real numbers,

i.e. 0 and 0.

Think and as small positive numbers.

“x be sufficiently close to but different from a”:

0 x a where a x a x a

Limit of a Function

Let be a function defined on some open interval

containing , except possibly at .

The of the function as approaches is ,

written as lim ,

if for every

lim

0, there exists a 0,

it

x a

f

a a

x a L

f x L

such that

if 0 , then .x a f x L

Example

2

0

2

Given 4 , find the following limits.

1. lim 4

2. lim 0

x

x

f x x

f x

f x

-3 -2 -1 1 2 3

-1

1

2

3

4

5

x

y

Proving Limits of a Function

To prove that lim .

i. Verify that is defined on some open interval

containing , except possibly at .

ii. Consider a positive epsilon ( 0).

iii. Find an expression for such that

whene

x af x L

f

a a

f x L

ver 0 .

iv. Show that if 0 then .

v. Form a conclusion.

x a

x a f x L

Example

3Prove using the definition that lim 2 1 5.

Preliminary Analysis.

We want such that if then .

2 1 5 2 3

2 6 32

2 3 Choose .2

xx

x a f x L

x x

x x

x

Proof.

i. Since the function is a polynomial function,

it is defined on any open interval containing 3.

ii. Let 0.

iii. Choose .2

iv. If 0 3 , then 32

2 3

2 6

2 1 5

x x

x

x

x

3

v. Therefore, lim 2 1 5.x

x

Example

1Prove using the definition that lim 4 3 7.

Preliminary Analysis.

We want such that if then .

4 3 7 3 1

3 3 13

3 1 Choose .3

3 1

xx

x a f x L

x x

x x

x

x

Proof.

i. Since the function is a polynomial function,

it is defined on any open interval containing 1.

ii. Let 0.

iii. Choose .3

iv. If 0 1 , then 13

3 1

3 1

3 3

x x

x

x

x

1

3 3

4 3 7 .

v. Therefore, lim 4 3 7.x

x

x

x

Challenge!

Given 0, prove that lim .x a

m mx b ma b