Section 1.4 – Continuity and One-Sided Limits

Section 1.4 – Continuity and One-Sided Limits

ExampleFind values of a and b that makes f(x) continuous.

f x ax 3 if x 58 if x 5

x 2 +bx +1 if x 5

When x=5, all three pieces must have a limit of 8.

5a 5

ax 3 8

a 5 3 8

a 1

5b 28 8

x 2 bx 3 8

5 2 b 5 3 8

5b 20

b 4

Continuity at a Point

A function f is continuous at c if the following three conditions are met:

1. is defined.

2. exists.

3.

( )f c

lim ( )x c

f x

lim ( ) ( )x c

f x f c

c

L

f(x)

x

For every question of this type, you need (1), (2), (3), conclusion.

Example 1

Show is continuous at x = 0.

f x 2 1 x 2

1. f 0

2 1 02

1The function is clearly

defined at x= 0

2. limx 0

2 1 x 2

2 1 02

1With direct

substitution the limit clearly exists at x=0

3. f 0 limx 0

2 1 x 2The value of the

function clearly equals the limit at x=0

f is continuous at x = 0

Example 2

Show is not continuous at x = 2. 8 1 if 210 if 2

x xf x

x

1. 2f 10 The function is clearly 10 at x = 2

2

lim 8 1x

x

8 2 1 15

With direct substitution the limit clearly exists at x=0

2

3. 2 lim x

f f x

The value of the function clearly does not equal the limit at x=2

f is not continuous at x = 2

The behavior as x approaches 2 is dictated by 8x-1

2

2. lim x

f x

DiscontinuityIf f is not continuous at a, we say f is discontinuous

at a, or f has a discontinuity at a.

Typically a hole in the curve

Step/Gap Asymptote

Types Of DiscontinuitiesRemovable

Able to remove the “hole” by defining f at one point

Non-RemovableNOT able to remove the “hole” by defining f at

one point

ExampleFind the x-value(s) at which is not

continuous. Which of the discontinuities are removable?

22 153( ) x x

xf x

There is a discontinuity at x=-3 because this makes the

denominator zero.

If f can be reduced, then the discontinuity is removable:

2 5 33

x xxf x

2 5 33

x xx

2 5x This is the

same function as f except at

x=-3f has a removable

discontinuity at x = -3

Notice that:

003f

Indeterminate Form: 0/0

Let:

g xf x

h x

If:

00

g cf c

h c

Then f(x) has a removable discontinuity at x=c.

If f(x) has a removable discontinuity at x=c.

Then the limit of f(x) at x=c exists.

c

L

f(x)

x

One-Sided Limits: Left-HandIf f(x) becomes arbitrarily close to a single REAL

number L as x approaches c from values less than c, the left-hand limit is L.

The limit of f(x)…

as x approaches c from the left…

is L.Notation:

c

L

f(x)

x

lim ( )x c

f x L

lim ( )x c

f x L

One-Sided Limits: Right-HandIf f(x) becomes arbitrarily close to a single REAL

number L as x approaches c from values greater than c, the right-hand limit is L.

The limit of f(x)…

as x approaches c from the right…

is L.Notation:

c

L

f(x)

x

Example 1Evaluate the following limits for

f x x

limx 1

f x

limx 1

f x

limx 1

f x

1

0

DNE

3.5

limx

f x

3.5

limx

f x

3.5

limx

f x

3

3

3

Example 2

Analytically find . 12

24

3 if 4lim if

if 4x

x xf x f x

x x

12If is approaching 4 from the left, the function is defined by 3x x

4

limx

f x

124

lim 3x

x 1

2 4 3

4

Therefore lim 5x

f x

5

The Existence of a Limit

Let f be a function and let c be real numbers. The limit of f(x) as x approaches c is L if and only if

lim ( ) lim ( )x c x c

f x L f x

c

L

f(x)

x

Right-Hand LimitLeft-Hand Limit =A limit exists if…

Example 1Analytically show that .

limx 2

x 2 1 1

Evaluate the right hand limit at 2 :x

limx 2

x 2 1

limx 2

x 2 1

2 2 1

2 1

x

if 2if 2

xx

1

Evaluate the left hand limit at 2 :x

limx 2

x 2 1

limx 2

x 2 1

2 2 1

1

Therefore limx 2

x 2 1 1

Use when x>2

Use when x<2You must use the piecewise equation:

2 +12 +1

xx

Example2Analytically show that is continuous at x = -

1. = 1f x x

Evaluate the right hand limit at 1:x

1lim 1

xx

1lim 1

xx

1 1

f x

if 1if 1

xx

0

Evaluate the left hand limit at 1:x

1lim 1

xx

1lim 1

xx

1 1 0

Therefore is continuous at 1f x x

Use when x>-1

Use when x<-1

You must use the piecewise equation:

11

xx

1. 1f 1 1 0

1

2. Find limx

f x

0

1

3. 1 lim x

f f x

Continuity on a Closed Interval

A function f is continuous on [a, b] if it is continuous on (a, b) and

lim ( ) ( ) lim ( ) ( )x a x b

f x f a and f x f b

a

f(a)

f(b)

xb

Must have closed dots on the endpoints.

t x

Example 1Use the graph of t(x) to determine the intervals on

which the function is continuous.

6, 3 3,0 0,2 2,5 5,6

Example 2Discuss the continuity of

f x 1 1 x 2

The domain of f is [-1,1]. From our limit properties, we can say it is continuous on (-

1,1)

By direct substitution:

limx 1

f x

1 1 1 2

1

f 1

limx 1

f x

1 1 1 2

1

f 1

Is the middle is continuous?

Are the one-sided limits of the endpoints equal to the functional value?

f is continuous on [-1,1]

Properties of ContinuityIf b is a real number and f and g are continuous at x = c,

then following functions are also continuous at c:1. Scalar Multiple:

2. Sum/Difference:

3. Product:

4. Quotient: if

5. Composition:

Example: Since are continuous, is continuous too.

f og x

f x g x

f x g x

b f x

f x g x

g c 0

f x 2x and g x x 2

h x 2x x 2

Intermediate Value TheoremIf f is continuous on the closed interval [a, b] and k

is any number between f(a) and f(b), then there is at least one number c in [a, b] such that:

( )f c k

a b

f(a)

f(b)

k

c

This theorem does NOT find

the value of c. It just proves it

exists.

Free Response Exam 2007

1h 1 6f g 2 6f 9 6 3

3h 3 6f g 4 6f 1 6 7

Since h(3) < -5 < h(1) and h is continuous, by the IVT, there

exists a value r, 1 < r < 3, such that h(r) = -5.

Notice how every part of the theorem is

discussed (values of the function AND continuity).

We will learn later that this implies continuity.

ExampleUse the intermediate value theorem to show

has at least one root.

f x 4 x 3 6x 2 3x 2

f 0 4 0 3 6 0 2 3 0 2

2

f 2 4 2 3 6 2 2 3 2 2

12Find an output greater than zero

Find an output less than zero

Since f(0) < 0 and f(2) > 0

There must be some c such that f(c) = 0 by the IVTThe IVT can be used since f

is continuous on [-∞,∞].

ExampleShow that has at least one solution on the

interval .

cos x x 3 x

f 4 cos

4 4 3

4

1.008

f 2 cos

2 2 3

2

2.305Find an output less than zero

Find an output greater than zero

Since and

There must be some c such that cos(c) = c3 - c by the IVT

The IVT can be used since the left and right side are

both continuous on [-∞,∞].

4 ,

2 Solve the equation for zero.

cos x x 3 x 0

f x

f 4 0

f 2 0