Section 1.4 – Continuity and One-Sided Limits - Katy …staff.katyisd.org/sites/thscalculusap/20142015 BC Lessons/Unit 1... · Section 1.4 –Continuity and One-Sided Limits. Welcome

Section 1.4 – Continuity and

One-Sided Limits

Welcome to BC Calculus

Thursday Aug 28

Tonight’s HW: STUDY FOR QUIZ

2

3 if 5

8 if 5

+ +1 if 5

ax x

f x x

x bx x

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

List 3 conditions for CONTINUITY

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() ()xcfx fc

c

L

f(x)

x

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

Example

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

f x

ax 3 if x 5

8 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 38

a 5 3 8

a 1

5b28 8

x2 bx 3 8

5 2b 5 3 8

5b 20

b 4

Homework Question?

Make sure you have gone through 1.1 – 1.5 by Tuesday!

Required Discussion: Unit 1, What I did this Summer

Optional Discussion: Unit 0

UT Online 0 Due TONIGHT

What to do if we have a Fire Drill during a Test/Quiz…..

Properties of Continuity

If 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:

3. Quotient: if

4. Composition:

Example: Since are continuous,

is continuous too.

fogx

fxgx

fxgx

b f x

f x

g x

gc0

fx2x and gxx2

hx2xx2

Example 1

Show is continuous at x = 0.

fx21x2

1. f0

2102

1The function is clearly

defined at x= 0

2. limx0 21x2

2102

1With direct

substitution the limit

clearly exists at x=0

3. f0limx0 21x2

The 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 2

10 if 2

x xf x

x

1. 2f 10The 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

Discontinuity

If 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 Discontinuities

RemovableAble to remove the “hole” by

defining f at one point

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

one point

Example

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

continuous. Which of the discontinuities are removable?

22 153

() xxx

fx

There is a discontinuity at x=-3

because this makes the

denominator zero.

If f can be reduced, then the discontinuity

is removable:

25 3

3

x x

xfx

2 5 3

3

x x

x

2 5x

This is the

same function

as f except at

x=-3

f has a removable

discontinuity at x = -3

Notice that:

00

3f

Indeterminate Form: 0/0

Let:

g xf x

h x

If:

0

0

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-Hand

If 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()xc

fx L

lim()xc

fx L

One-Sided Limits: Right-Hand

If 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 1

Evaluate the following limits for

fxx

limx1fx

limx1fx

limx1fx

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

12

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

4

limx

f x

12

4lim 3

xx

1

24 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()xc xcfxLfx

c

L

f(x)

x

Right-Hand LimitLeft-Hand Limit =

A limit exists if…

Example 1

Analytically show that .

limx2

x 2 11

Evaluate the right hand limit at 2 :x

limx2

x 2 1

limx2

x 2 1

2 2 1

2 1

x

if 2

if 2

x

x

1

Evaluate the left hand limit at 2:x

limx2

x 2 1

limx2

x 2 1

2 2 1

1

Therefore limx2x 2 11

Use

when

x>2

Use

when

x<2You must use the piecewise equation:

2 +1

2 +1

x

x

Example2

Analytically 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 1

if 1

x

x

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:

1

1

x

x

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()()xa xbfxfaandfxfb

a

f(a)

f(b)

x

b

Must have closed dots on the endpoints.

t x

Example 1

Use 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 2

Discuss the continuity of

fx11x2

The domain of f is [-1,1]. From our limit

properties, we can say it is continuous on (-

1,1)

By direct substitution:

limx1fx

1112

1

f1

limx1fx

1112

1

f1

Is the middle is continuous?

Are the one-sided limits of the endpoints

equal to the functional value?

f is continuous on [-1,1]

Intermediate Value Theorem

If 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:

()fc 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.

Example

Use the intermediate value theorem to show

has at least one root.

fx4x36x23x2

f0403602302

2

f2423622322

12

Find 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 [-∞,∞].

Example

Show that has at least one solution on the

interval .

cos x x3 x

f 4 cos

4 4 3

4

1.008

f 2 cos

2 2 3

2

2.305

Find 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 x3 x 0

f x

f 4 0

f 2 0

Squeeze Theorem

If ( ) ( ) ( ) in some deleted neighborhood of

and lim ( ) lim ( )

then lim ( )

x a x a

x a

f x g x h x a

f x h x L

g x L

Squeeze Theorem

( ) ( ) ( )f x g x h x

x

y

a

)(xf

)(xg

)(xh

Squeeze Theorem

x

y

L

a

)(xf

)(xh

( ) ( ) ( )f x g x h x

lim ( ) lim ( )x a x a

f x h x L

Squeeze Theorem

x

y

L

a

)(xf

)(xg

)(xh

( ) ( ) ( )f x g x h x

lim ( ) lim ( )x a x a

f x h x L

lim ( )x a

g x L

Squeeze Theorem

You will see

this type of

idea over and

over again.

x

y

L

a

)(xf

)(xg

)(xh

2 2

0 0 0

1 1lim sin lim limsinx x x

x xx x

Example 1

Example 1

2 2

0 0 0

1 1lim sin lim limsinx x x

x xx x

Example 1

• We cannot apply the limit laws since

DNE (2.1.1)

xx

1sinlim

0

2 2

0 0 0

1 1lim sin lim limsinx x x

x xx x

Example 1

( ) ( ) ( )f x g x h x

lim ( ) lim ( )x a x a

f x h x L

lim ( )x a

g x L

Make sure to quote the name of the Squeeze

Theorem.

1sin

x