introduction to calculus set

8/8/2019 Introduction to Calculus Set

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Introduction to Calculus

Bhaskara Aacharya (1114 1185)

Legendary Indian Mathematician who gave preliminary concepts of infinitesimal

calculus and mathematical analysis , in his work , Siddhanta Shiromani.


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Calculus is the mathematics of change.

It has two major branches, differential calculus and integral calculus, which arerelated by the Fundamental Theorem of Calculus.

Differential calculus determines varying rates of change. It helps solve

problems involving force

Integration is the "inverse" (or opposite) of differentiation or a sum of series. It

measures accumulations over periods of change. Integration can find volumes

and lengths of curves, measure forces and work, etc.

Calculus has widespread applications in science, economics, finance and

engineering.


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2.3 Introduction to Functions

A is any set of ordered pairs.The set of all first components of the

Definition of a Relation

relation

domainordered pairs is called the of

the relation, and the set of all second

components is called the of the

relat

range

ion.


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Ex 1: Determine whether each

relation is a function.

a. {(4,5), (6,7), (8,8)}

b. {(5,6), (4,7), (6,6), (6,7)}

We begin by making a igure

or each relation that sho s set , the

domain, and set , t

So

he

luti

ra

o

e.

n

ng

X

Y


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Solution for part (a)

4

6

8

5

7

8

X Y

Domain ange

The figure shows that every

element in the domaincorresponds to exactly one

element in the range.

No two ordered pairs in the given relation have

the same first component different second

components. Thus, the relation is a function


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Solution for part (b)

4

5

6

6

7

X Y

Domain ange

The igure sho s that 6corresponds to both 6 and 7.

I any element in the domain corresponds tomore than one element in the range, the

relation is not a unction, Thus, the relation

is not a unction.


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Function Notation

( )

The variable is called the, because it can be assigned any

of the permissible numbers from the domain.

The variable is called the

independentvariable

dependent

, becausevar itsiable value depe

y f x

x

y

!

nds on .x


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Function Notation

The special notation ( ), read " of "

or " at ," represents the value of the

function at the number .

The notation ( ) does not mean

" times ."

f x f x

f x

x

f x

f x


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Ex2: Determine whether each equation

defines yas a function ofx.

2

1. 25

2. 25

x y

x y


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Solution continued

From this last equation we can see that

for each value of , there is one and

only one value of . Thus, the equation

defines y as a function of .

x

y

x


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Solution of22

25x y !2 25y x!

25y x! s The in this last equation sho s that

there are t o values o or in thedomain o

i.e the interval ( ,5). For this reason

the equation does not de ine as a unction o .

y x

f

y x

s

g


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Finding a Functions Domain

I a unction does not model data or

verbal conditions, its domain is the

largest set o real numbers or hich the

value o ( ) is a real number.

f

f x


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Finding a Functions Domain

Exclude rom a unction's domain real

numbers that cause division by zero

and real numbers that result in an even

root o a negative number.


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Ex 5: Find the domain of each

function2a. ( ) 8 5 2

2 b. ( )

5

c. ( ) 2

f x x x

g xx

h x x

!

!

!


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Solution part a

2The unction ( ) 8 5 2 contains

neither division nor an even root.

The domain o is the set o all real numbers.

f x x x

f

!


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Solution part b

2

The function ( ) contains division.5

Because division by zero is undefined, we

must exclude from the domain values of

that cause 5 to be 0. Thus cannot equal

to 5. The domain of function is

g x x

x

x x

g

!

{ | 5}.x x {


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Solution part c

The function ( ) 2 contains an even

root. Because only non-negative numbershave real square roots, the quantity under the

radical sign, 2 must be greater than or

equal to 0. Thus, 2 0 or 2

Th

h x x

x

x x

!

u u

erefore the domain of is { | 2}

or the interval [ 2, ).

h x x u

g


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Limit

We say that the limit o ( ) as approaches is and rite f x x a L

lim ( )x a

f x Lp

!

if the values of ( ) approach as approaches . f x x a

a

( ) y f x!


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c) Find2

3 if 2lim ( ) where ( )

1 if 2x

x x f x f x

xp

{ !

!

-2

62 2

lim ( ) = lim 3

x x

f x xp

p

Note: f(-2) 1

is not involved

23 lim

3( 2) 6

xx

p!

! !


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The Definition ofLimit-I H

lim ( )We say if and only ifx a

f xp

!

given a positive number , there exists a positive such thatI H

if 0 | | , then | ( ) | . x a f xH I

( ) y f x!a

LL I

L I

a H a H


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such that or all in ( , ), x a a aH H{

then we can find a (small) interval ( , )a aH H

( ) is in ( , ). f x L LI I

This means that i e are given a

small interval ( , ) centered at , L L LI I


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Examples2

1. Sho that lim(3 4) 10.x

xp

!

Let 0 be given.I " We need to find a 0 such thatH "if - 2 | ,x H then | (3 4) 10 | .x I

But | (3 4) 10 | | 3 6 | 3 | 2 | x x x I ! !

i | 2 |3

x I So we choose .3

IH !

1

12. Show that lim 1.

x xp!

Let 0 be given. We need to find a 0 such thatI H" "1if | 1 | , then | 1 | .xx

H I

1 11But | 1 | | | | 1 | .x

xx x x

! ! What do we do with the

x?


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1 31I e decide | 1| , then .2 22

x x

1And so


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The right-hand limit off(x), asx approaches a,

equals L

written:

if we can make the valuef(x) arbitrarily close

to Lby takingx to be sufficiently close to the

right ofa.

lim ( )x a

f x Lp

!

a

L

( ) y f x!

One-Sided LimitOne-Sided Limits


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The left-hand limit off(x), asx approaches a,

equals M

written:

if we can make the valuef(x) arbitrarily close

to Lby takingx to be sufficiently close to theleft ofa.

lim ( )x a

f x Mp

!

a

M

( ) y f x!


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2 i 3( )

2 i 3

x xf x

x x

e! "

1. Given

3

lim ( )x

f x

p

3 3lim ( ) lim 2 6

x x f x x

p p! !

2

3 3lim ( ) lim 9

x x f x x

p p! !

Find

Find3

lim ( )x

f xp

Examples

Examples ofOne-Sided Limit


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1, i 0

2. Let ( ) 1, i 0.

x x

f x x x

"

! e Find the limits:

0lim( 1)

xx

p! 0 1 1! !

0a) lim ( )

xf x

p

0 b) lim ( )

xf x

p 0lim( 1)

xx

p! 0 1 1! !

1

c) lim ( )x

f xp 1

lim( 1)x

xp

! 1 1 2! !

1d) lim ( )

xf x

p 1lim( 1)

xx

p! 1 1 2! !

More Examples


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lim ( ) i and only i lim ( ) and lim ( ) .x a x a x a f x L f x L f x L p p p! ! !

For the function

1 1 1lim ( ) 2 because lim ( ) 2 and lim ( ) 2.x x x

f x f x f x p p p

! ! !But

0 0 0lim ( ) does not exist because lim ( ) 1 and lim ( ) 1.x x x f x f x f x p p p! !

1, i 0( )

1, i 0.

x xf x

x x

"!

e

This theorem is used to show a limit does not

exist.

ATheorem


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Examples Using Limit Rule

Ex.

2

3lim 1x xp

2

3 3lim lim1x xxp p!

2

3 3

2

lim lim1

3 1 10

x xx

p p!

! !

Ex.1

2 1lim

3 5x

x

xp

1

1

lim 2 1

lim 3 5

x

x

x

x

p

p

!

1 1

1 1

2 lim lim1

3lim lim5

x x

x x

x

x

p p

p p

!

2 1 1

3 5 8

! !


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More Examples

3 31. Suppose lim ( ) 4 and lim ( ) 2. Findx x f x g xp p! !

3

a) lim ( ) ( )x

f x g xp

3 3

lim ( ) lim ( )x x

f x g xp p

!

4 ( 2) 2! !

3

b) lim ( ) ( )x

f x g xp

3 3

lim ( ) lim ( )x x

f x g xp p

!

4 ( 2) 6! !

3

2 ( ) ( )c) lim

( ) ( )x

f x g x

f x g xp

3 3

3 3

lim 2 ( ) lim ( )

lim ( ) lim ( )

x x

x x

f x g x

f x g x

p p

p p

!

2 4 ( 2) 5

4 ( 2) 4

! !


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Indeterminate forms occur when substitution in the limit

results in 0/0. In such cases either factor or rationalize the

expressions.

Ex.25

5lim

25x

x

xp

Notice form

5

5lim

5 5x

x

x xp

!

Factor and cancel

common factors

51 1

lim5 10x xp

! !

Indeterminate Forms


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The Squeezing Theorem

I ( ) ( ) ( ) hen is near , and if x g x h x x ae e, thenlim ( ) lim ( )

x a x a f x h x L

p p! ! lim ( )

x ag x L

p!

2

0

Show that liExampl m 0e: .x

x sin

x

T

p

!

0Note that e cannot use product rule because limx sin xTp DNE! But 1 sin 1xT e e 2 2 2and so sin . x x xxT e e

2 2

0 0Since lim lim( ) 0,x xx xp p! ! we use the Squeezing Theorem to conclude

20

lim 0.x

x sinx

T

p!

See Graph


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Continuity

A functionfis continuous at the pointx =a if

the following are true:

) ( ) is definedi f a

) lim ( ) existsx a

ii f xp

a

f(a)


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A functionfis continuous at the pointx =a if

the following are true:

) ( ) is definedi f a

) lim ( ) existsx a

ii f xp

) lim ( ) ( )x a

iii f x f ap

!

a

f(a)

xamp e


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At which value(s) of x is the given function

discontinuous?

1. ( ) 2 f x x! 2

92 . ( )3

xg xx

!

Continuous everywhereContinuous everywhere

except at 3x !

( 3) is undefinedg

lim( 2) 2x a x ap !

and so lim ( ) ( )x a

f x f ap

!

-4 -2 2 4

-2

2

4

6

-6 -4 -2 2 4

-10

-8

-6

-4

-2

2

4

xamp e

s


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2, if 13. ( )

1, if 1

x xh x

x

"!

e

1lim ( )

xh x

pand

Thus h is not cont. atx=1.

1!1

lim ( )x

h xp

3!

h is continuous everywhere else

1, if 04. ( )

1, if 0

xF x

x

e!

"

0lim ( )x F xp 1! and 0lim ( )x xp 1!

ThusFis not cont. at 0.x !

F is continuous everywhere else

-2 2 4

-3

-2

-1

1

2

3

4

5

-1

-5 5 1

-3

-2

-1

1

2

3


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Continuous Functions

Apolynomial function y =P(x) is continuous at

every pointx.

A rational function is continuous

at every pointx in its domain.

( )( )

( )p x

R xq x

!

Iffandgare continuous atx =a, then

, , and ( ) 0 are continuous

at

f f g fg g a

g

x a

s {

!


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Example

? A

2Given ( ) 3 2 5,

Show that ( ) 0 has a solution on 1, 2 .

f x x x

f x

!

!

(1) 4 0

(2) 3 0

f

f

! ! "

f(x) is continuous (polynomial) and sincef(1) < 0andf(2) > 0, by the Intermediate Value Theorem

there exists a c on [1, 2] such thatf(c) = 0.


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Limits at Infinity

For all n > 0,1 1lim lim 0

n nx xx xpg pg! !

provided that is defined.

1

nx

Ex.2

2

3 5 1lim

2 4x

x x

xpg

2

2

5 13lim

2 4x

x x

xpg

!

3 0 0 3

0 4 4

! !

Divide

by2

x

2

2

5 1lim 3 lim lim

2lim lim 4

x x x

x x

x x

x

pg pg pg

pg pg

!


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More Examples

3 2

3 2

2 3 21. lim 100 1x

x x

x x xp g

3 2

3 3 3

3 2

3 3 3 3

2 3 2

lim100 1x

x x

x x x x x x

x x x x

pg

!

3

2 3

3 22

lim1 100 1

1x

x x

x x x

p g

!

22

1! !


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0!

2

3 2

4 5 212. lim

7 5 10 1x

x x

x x xpg

2

3 3 3

3 2

3 3 3 3

4 5 21

lim7 5 10 1x

x x

x x x

x x x

x x x x

pg

!

2 3

2 3

4 5 2 1

lim5 10 1

7x

x x x

x x x

p g

!

07

!

2 2 43. lim

12 31x

x x

xpg

2 2 4

lim12 31x

x x

x x xx

x x

pg

!

42

lim31

12x

xx

x

pg

!

21 2

g !

! g


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24. lim 1x

x xpg

2 22

1 1lim

1 1x

x x x x

x xpg

!

2 2

2

1lim

1x

x x

x xpg

!

2

1lim

1x x xpg

! 1 1

0! ! !

g

g g


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Infinite LimitsFor all n > 0,

1

limn

x a x ap

! g

1

lim if is evenn

x an

x ap

! g

1

lim i is oddn

x an

x ap

! g

-8 -6 -4 -2 2

-20

-15

-10

-5

5

10

15

20

-2 2 4 6

-20

-10

10

20

30

40

ore Graphs

-8 -6 -4 -2 2

-20

-15

-10

-5

5

10

15

20


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Examples

Find the limits2

20

3 2 11. lim

2x

x x

xp

2

0

2 13= lim

2x

x xp

3

2

g g! ! g

3

2 12. lim

2 6x

x

xp

3

2 1= lim

2( 3)x

x

xp

! g

-8 -6 -4 -2 2

-20

20

40


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Limit and Trig Functions

From the graph of trigs functions

( ) sin and ( ) cos f x x g x x! !

we conclude that they are continuous everywhere

-10 -5 5 10

-1

-0.5

0.5

1

-10 -5 5 10

-1

-0.5

0.5

1

limsin sin and lim cos cos x c x c

x c x cp p

! !


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Examples

2a) lim sec

x

xT

p

! g 2

b) lim secx

xT

p

! g

3 2c) lim tanx xTp

! 3 2

d) lim tanx

xT p

!

e) lim cotx

xT

p! g

g

3 2g) lim cot

x

xTp

! 3 2

cos 0lim 0

sin 1x

x

xTp! !

4

) lim tanx

xTp

!g

1

Li it d E ti l


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Limit and Exponential

Functions

-6 -4 -2 2 4 6

-2

2

4

6

8

10

, 1x

y a a! "

-6 -4 -2 2 4 6

-2

2

4

6

8

10 , 0 1xy a a!

The above graph confirm that exponential

functions are continuous everywhere.

lim x cx c

a ap

!


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Asymptotes

horizontal asymptotThe line is called a

of the curve ( ) if eihter

ey L

y f x

!

!

lim ( ) or lim ( ) .x x

f x L f x Lpg pg! !

vertical asymptoteThe line is called a

o the curve ( ) i eihter

x c

y f x

!

!

lim ( ) or lim ( ) . x c x c

f x f x p p

! sg ! sg


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Examples

Find the asymptotes of the graphs of the functions

2

2

11. ( )

1

xf x

x

!

1(i) lim ( )

xf x

p! g

There ore the line 1

is a vertical asymptote.

x !

1.(iii) lim ( )x

f xpg

!

1(ii) lim ( )

xf x

p! .g

Therefore the line 1


x !


is a horizonatl asymptote.

y !

-4 -2 2 4

-10

-7.5

-5

-2.5

2.5

5

7.5

10


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2

12. ( )

1

xf x

x

!

21 1

1(i) lim ( ) lim

1x x

xf x

xp p !

1 1

1 1 1= lim lim .

( 1)( 1) 1 2x x

x

x x xp p

! !

There ore the line 1

is a vertical asympNO t eT ot .

x !

1(ii) lim ( ) .

xf xp ! g



x !

(iii) lim ( ) 0.x

f xpg

!


is a horizonatl asymptote.

y !

-4 -2 2 4

-10

-7.5

-5

-2.5

2.5

5

7.5

10


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introduction to calculus set

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