State Standard – 1.0 Students demonstrate knowledge of limit of values of functions. This includes one-sided limits, infinite limits, and limits at infinity.

State Standard – 1.0 Students demonstrate knowledge of limit of values of functions. This includes one-sided limits, infinite limits, and limits at infinity.

Objective – To be able to find the limit of a function.

Definition:

And say “the limit of f(x), as x approaches a, equals L”

This says that the values of f(x) get closer and closer to the number L as x gets closer to the number a (from either side)

lim ( )x a

f x L

Example 1

Find the Limit of f(x) for these three cases:

a a rnn

11

3

3

3

) lim ( )

) lim ( )

) lim ( )

x

x

x

a f x

b f x

c f x

2

2

2

Example 2

Find the Limit of f(x) for these four cases:

3

3

3

) lim ( )

) lim ( )

) lim ( )

) (3)

x

x

x

a f x

b f x

c f x

d f

2

2

2

DNE

Example 3

Find the Limit of f(x) for these four cases:

2

2

2

) lim ( )

) lim ( )

) lim ( )

) (2)

x

x

x

a f x

b f x

c f x

d f

1

3

DNE

1.5

–5 –4 –3 –2 –1 1 2 543

–5

–4

–3

–2

–1

1

2

5

4

3

Example 4


- ∞

+ ∞

DNE

2

2

2

) lim ( )

) lim ( )

) lim ( )

x

x

x

a f x

b f x

c f x

Example 5


0

0

0

) lim ( )

) lim ( )

) lim ( )

x x

x x

x x

a f x

b f x

c f x

+ ∞

+ ∞

+ ∞

Example 6

Find the Limit of f(x) for these two cases:

) lim ( )

) lim ( )

x

x

a f x

b f x

-1

4

Example 7

Find the Limit of f(x) for these two cases:

) lim ( )

) lim ( )

x

x

a f x

b f x

+ ∞

-2

WS on Limits

Pg. 102

4 – 9



Example 1

a a rnn

11

+ ∞

+ ∞

0

+ ∞

DNE

-1

Example 2

– ∞

– ∞

2

– ∞

1

2

Example 3

0

0

DNE

0

0

DNE

–5 –4 –3 –2 –1 1 2 543

–5

–4

–3

–2

–1

1

2

5

4

3

DNE

0

+ ∞

DNE

0

– ∞



Example 1

Use a table of values to estimate the value of the limit.

a a rnn

11

0.75 0.90 0.99 ?0.999 1.001 1.01 1.1 1.25

x approaches 1

from the LEFT

x approaches 1

From the RIGHT

= 1

-0.1 -0.05 -0.001 0 0.001 0.05 0.1

f(x)

x

Example 2


a a rnn

11

100 400 1x106 ? 1x106 400 100

x approaches +∞

from the LEFT

x approaches +∞

From the RIGHT

= +∞20

1limx x

1.95 1.995 1.999 2 2.001 2.015 2.1

f(x)

x

Example 3


a a rnn

11

63.20 63.92 63.998 ? 64.001 64.24 65

x approaches 64

from the LEFT

x approaches 64

From the RIGHT

2

2

16 64lim

2x

x

x

= 64

Pg. 103

15 – 20

-1.1 -1.05 -1.001 -1 -0.999 -0.95 -0.8

f(x)

x


a a rnn

11

-9 -19 -999 ? 1001 21 6

x approaches –∞

from the LEFT

x approaches +∞

From the RIGHT

= DNE1

2lim

1x

x

x



Example for case 1:

Ex for case 2:= 16

0lim16x

= 85

lim8x = 2

2limx

x

= -44

limx

x

Ex for case 3:

= 322

3limx

x

= 9

Limit Laws Suppose that c is a constant:

lim ( )x a

f x

and lim ( )x a

g x

lim[ ( ) ( )] lim ( ) lim ( )x a x a x a

f x g x f x g x

Sum and Difference:

lim[ ( ) ( )] lim ( ) lim ( )x a x a x a

f x g x f x g x

Product:

lim ( )( )lim

( ) lim ( )x a

x ax a

f xf x

g x g x

Division:

lim[ ( )] lim ( )x a x a

cf x c f x

Scalar Mult.:

Example 1a

Find the Limit:

2

2lim 3 5 4x

x x

2

2 2 2lim3 lim 5 lim 4x x x

x x

2

2 2 23lim 5lim lim 4

x x xx x

23 2 5(2) 4

3(4) – 10 + 4 = 6

Example 1b

Find the Limit: 3

23

5 4lim

2x

x x

x

3

2

3 5(3) 4

3 2

27 15 4

9 2

16

7

Example 2

Find the Limit:

( ) 7f x x

3) lim ( )

xa f x

3 7

4

2( )g x x

4) lim ( )

xb g x

24

16

3) lim ( ( ))

xc g f x

( 3 7)g

16

2(4)

Pg. 111-112

1 – 9

Evaluate the limit.

= 3/2

4

2 1) lim

3x

xb

x

1

2) lim

1x

xa

x

= 9



Example 1

Find:

Functions with the Direct Substitution Property are called continuous at a. However, not all limits can be evaluated by direct substitution, as the following example shows:

2

1

1lim

1x

x

x

1

( 1)( 1)lim

1x

x x

x

= 1+11

lim 1x

x

= 2

Example 2

Find the Limit:

22

2lim

4x

x

x

2

1( 2)lim

( 2)( 2)x

x

x x

2

1lim

2x x

1

22

1

4

Example 3

Find the Limit: 2

24

5 4lim

2 8x

x x

x x

4

( 4)( 1)lim

( 4)( 2)x

x x

x x

4

1lim

2x

x

x

4 1

4 2

1

2

3

6

0lim

( 2 2)x

x

x x

Example 4

Find the Limit:

0

2 2limx

x

x

2 2

2 2

x

x

0

2 2lim

( 2 2)x

x

x x

2

40

1lim

2 0 2x 1

2 2

1

2 2

2

2

0lim

( 5 5)x

x

x x

Example 5

Find the Limit:

0

5 5limx

x

x

5 5

5 5

x

x

0

5 5lim

( 5 5)x

x

x x

5

100

1lim

0 5 5x 1

5 5

1

2 5

5

5

Pg. 112

11 – 29 odd

Evaluate the limit.

= DNE

2

4

2 32) lim

4x

xb

x

2

1

2 1) lim

1x

x xa

x

= 16



If a function f is not continuous at a point c, we say that f is discontinuous at c or c is a point of discontinuity of f.

Most of the techniques of calculus require that functions be continuous. A function is continuous if you can draw it in one motion without picking up your pencil.

A function is continuous at a point if the limit is the same as the value of the function.

This function has discontinuities at x=1 and x=2.

It is continuous at x=0, x=3, and x=4, because the one-sided limits match the value of the function

1 2 3 4

1

2

jump infinite oscillating

Essential Discontinuities:

Removable Discontinuities:

(You can fill the hole.)

Removing a discontinuity:

3

2

1

1

xf x

x

has a discontinuity at .1x

Write an extended function that is continuous at .1x

3

21

1lim

1x

x

x

2

1

1 1lim 1 1x

x x xx x

1 1 1

2

3

2

3

2

1, 1

13

, 12

xx

xf x

x

Note: There is another discontinuity at that can not be removed.

1x

Removing a discontinuity:

3

2

1, 1

13

, 12

xx

xf x

x

Note: There is another discontinuity at that can not be removed.

1x

Example

Find the value of x which f is not continuous, which of the discontinuities are removable?

2( )

xf x

x x

( 1)

x

x x

1

1x

0x Removable discontinuity is at:

Where as x – 1 is NOT a removable discontinuity.

Continuous functions can be added, subtracted, multiplied, divided and multiplied by a constant, and the new function remains continuous.

WS

1 – 10, 13 – 17 odd

Pg. 133

1-6, 10-12, and 15 – 20

–1–5 –4 –3 –2 –1 1 2 543

4

1

2

3

5

6

9

8

7

10Describe the continuity of the graph.



Definition: The line y = L is called a horizontal asymptote of the curve y = f(x) if either

orlim ( )x

f x L

lim ( )x

f x L

Case 1: Same Degree

2

2

6 2 1

5 3

x xy

x

Case 2: Degree smaller in Numerator

2

2 3

1

xy

x x

Case 3: Degree smaller in Denominator

22 3

7

xy

x

Example 1

Case 1: Numerator and Denominator of Same Degree

2

2

5 8 3lim

3 2x

x x

x

Divide numerator and denominator by x2

2

2

8 35

lim2

3x

x x

x

0 0

05 0 0

3 0

5

3

Example 2

Case 2: Degree of Numerator Less than Degree of Denominator

3

11 2lim

2 1x

x

x

Divide numerator and denominator by x3

2 3

3

11 2

lim1

2x

x x

x

0 0

00 0

2 0

0

2 0

Example 3

Case 3: Degree of Numerator Greater Than Degree of Denominator

22 3lim

7 4x

x

x

Divide numerator and denominator by x

32

lim4

7x

xx

x

0

0 2

lim7x

x

Example 4

a) b)3

3 2

4 3lim

8 2 1x

x

x x

3

2

4 7lim

2 3 10x

x x

x x

3

3

34

lim2 1

8x

x

x x

0

0 0

4

8

1

2

2

74

lim3 10

2x

xx

x x

0

0

4

lim2x

x

0

Example 5

Find the Limit:

22 1lim

3 5x

x

x

2/ x

2

12

lim5

3x

x

x

2

3

2/ x

/ x/ x

2 0lim

3 0x

Example 6

Find the Limit: 24 5lim

2 1x

x

x

2/ x

2

54

lim1

2x

x

x

21

2

2/ x

/ x/ x

4 0lim

2 0x

WS 1 – 8

and

Pg. 147

11 – 18, and 20 – 22

Solve and show work!

4 5

5

4 1

lim1

2x

x x

x

0 0

00 0

2 0

0

2 0

5

4 1lim

2 1x

x

x

State Standard – 4.1 Students demonstrate an understanding of the derivative of a function as the slope of the tangent line to the graph of the function.

Objective – To be able to find the tangent line.

Definition of a Tangent Line:

–1–5 –4 –3 –2 –1 1 2 543

4

1

2

3

5

6

9

8

7

10

PQ

Tan

gent

Lin

e

Slope:

a x

P(a,f(a))

Q (x,f(x))

x – a

f(x) – f(a)

( ) ( )PQ

f x f a

x am

Definition The tangent line to the curve y = f(x) at the point P(a,f(a)) is the line through P with the slope:

Provided that this limit exists.

( ) ( )limx a

f x f a

x am

Example 1

Find an equation of the tangent line to the parabola y=x2 at the point (2,4).

2

2

(2)lim

2x

x f

x

2 2 4

( ) ( )limx a

f x f a

x am

2

2

4lim

2x

x

x

2

( 2)( 2)lim

2x

x x

x

Use Point Slope

y – y1 = m (x – x1)

y – 4 = 4(x – 2)

y – 4 = 4x – 8

+4 +4

y = 4x – 4

Provided that this limit exists.

( ) ( )limx a

f x f a

x am

For many purposes it is desirable to rewrite this expression in an alternative form by letting:

h = x – a

Then x = a + h

0

( ) ( )limh

f a h f a

hm

0

( ) ( )limh

f a h f a

hm

Example 2

Find an equation of the tangent line to the hyperbola at the point (3,1).

3x

y

0

(3 ) (3)limh

f h f

hm

0

31

3limh

hh

0

3 33 3lim

h

hh h

h

0

3limh

hh

h

0lim

(3 )h

h

h h

0

1lim

3h h

1

3

y – y1 = m (x – x1)

y – 1 = -1/3(x – 3)

y – 1 = -1/3x + 1

+1 +1

y = -1/3x + 2

0

( ) ( )limh

f a h f a

hm

Example 3

Find an equation of the tangent line to the parabola y = x2 at the point (3,9).

0

(3 ) (3)limh

f h f

hm

2

0

(3 ) 9limh

h

h

0

(3 )(3 ) 9limh

h h

h

2

0

9 6 9limh

h h

h

2

0

6limh

h h

h

0lim 6h

h

6

y – y1 = m (x – x1)

y – 9 = 6(x – 3)

y – 9 = 6x – 18

+9 +9

y = 6x – 9

0

( ) ( )limh

f a h f a

hm

Example 4

Find an equation of the tangent line to the parabola y = x2–4 at the point (1,-3).

0

(1 ) (1)limh

f h f

hm

2 2

0

(1 ) 4 1 4limh

h

h

0

(1 )(1 ) 4 3limh

h h

h

2

0

1 2 1limh

h h

h

2

0

2limh

h h

h

0lim 2h

h

2

y – y1 = m (x – x1)

y – -3 = 2(x – 1)

y + 3 = 2x – 2

-3 -3

y = 2x – 5

Pg. 156

5a, 5b, 6a, 6b, 7 – 10, 11a, 12a, 13b, and 14b

State Standard – 4.0 Students demonstrate an understanding of the derivative of a function as the slope of the tangent line to the graph of the function.

Objective – To be able to find the derivative of a function.

limx c

f x f c

x c

is called the derivative of at .f c

We write: 0

limh

f a h f af a

h

“The derivative of f with respect to a is …”

There are many ways to write the derivative of y f x

f x “f prime x” or “the derivative of f with respect to x”

y “y prime”

dy

dx“dee why dee ecks” or “the derivative of y with

respect to x”

df

dx“dee eff dee ecks” or “the derivative of f with

respect to x”

df x

dx“dee dee ecks of eff of ecks” or “the derivative

of f of x”( of of )d dx f x

Example 1

Find the derivative of the function f(x) = x2 – 8x + 9 at the number ‘a’.

0

( ) ( )lim'( )h

f a h f a

hf a

2 2

0

( ) 8( ) 9 ( 8 9)limh

a h a h a a

h

2 2 2

0

2 8 8 9 8 9limh

a ah h a h a a

h

2

0

2 8limh

ah h h

h

0lim 2 8h

a h

2 8a

Example 2

Find the derivative of the function

0

( ) ( )lim'( )h

f x h f x

hf x

0

( ) 9 9limh

x h xx h x

h

2 2

0

( 9 9 ) ( 9 )( 9)( 9)

limh

x xh x h x xh xx x h

h

2 2

0

9 9 9lim

( 9)( 9)h

x xh x h x xh x

h x x h

0

9lim

( 9)( 9)h x x h

9( ) x

xf x

(x – 9) (x – 9)

(x +h – 9)

(x +h – 9)

9

( 9)( 9)x x

Pg. 163

13 – 17

State Standard – 1.0 Students demonstrate knowledge of limit of values of functions. This includes one-sided limits, infinite limits, and limits at infinity.

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