calculus i chapter two1 to a roman a “calculus” was a pebble used in counting and in gambling....

Calculus I Chapter two 1

To a Roman a “calculus” was a pebble used in counting and in gambling. Centuries later “calculare” meant” to compute,” “to figure out.” Today in mathematics and sciences calculus is elementary mathematics enhanced by the limit process.


Average Velocity Vs. Instantaneous Velocity

8Calculus I Chapter two

In the tables below we’ll find the average rate of change oftwo functions in three intervals.

( ) 2 1

2 1

1 3

2 5

3 7

4 9

f x x

x x

2

2

( ) 1

1

1 2

2 5

3 10

4 17

f x x

x x

(2) (1) 5 3

22 1 1

f f

(3) (2) 7 52

2 1 1

f f

(4) (3) 9 7

22 1 1

f f

(2) (1) 5 23

2 1 1

f f

(3) (2) 10 5

52 1 1

f f

(4) (3) 17 10

72 1 1

f f


( ) 2 1

2 1

1 3

2 5

3 7

4 9

f x x

x x

2

2

( ) 1

1

1 2

2 5

3 10

4 17

f x x

x x

(2) (1) 5 32

2 1 1

f f

(3) (2) 7 52

2 1 1

f f

(4) (3) 9 72

2 1 1

f f

(2) (1) 5 23

2 1 1

f f

(3) (2) 10 55

2 1 1

f f

(4) (3) 17 107

2 1 1

f f

The average rate of changeDOES NOT change! In factfor a line the average rate isthe same as instantaneous rate, and that is the slope!

For this function, the average rate of changeDOES change!


( ) 2 1

2 1

1 3

2 5

3 7

4 9

f x x

x x

2

2

( ) 1

1

1 2

2 5

3 10

4 17

f x x

x x

(2) (1) 5 32

2 1 1

f f

(3) (2) 7 52

2 1 1

f f

(4) (3) 9 72

2 1 1

f f

(2) (1) 5 23

2 1 1

f f

(3) (2) 10 55

2 1 1

f f

(4) (3) 17 107

2 1 1

f f

If we are asked to find the instantaneous rate at x=2, then we can say that: since the rate does not change the rate at x=2 is 2!

But, for this function the above question is a little tricky! Is the rate at x=2 3 or 5 or none? This is thequestion that was, finally,answered by Calculus!


2

2

( ) 1

1

1 2

2 5

3 10

4 17

f x x

x x

(2) (1) 5 23

2 1 1

f f

(3) (2) 10 55

2 1 1

f f

(4) (3) 17 107

2 1 1

f f

What if we had to answer the question without using calculus? Well, we can approximate the answer. But, how? We can say the rate isapproximately 3 or 5, but can we do better than that? Yes!

We use x values very close to 2, for example 2.1, 2.01, 2.001, or 2.0001! The closer the value to 2 the better the approximation!


:

( ) ( )

AverageVelocity from a to b

s b s a

b a


:

( ) ( )

AverageVelocity from a to b

s b s a

b a

Average velocities are approaching 0. So, we say the instantaneous velocity is 0 at π/2.


Geogebra file: TangentLine2

In order to compute the slope of the tangent line to the graph of y = f (x) at (a, f (a)), we compute the slope of the secant line over smaller and smaller time intervals of the form [a, x].


Thus we consider f (x)−f (a)/(x−a) and let x → a. If this quantity approaches a limit, then that limit is the slope of the tangent line to the curve y = f (x) at x = a.

Geogebra file: TangentLine2

( ) ( )?

f x f a

x awhen x a


Slope of the tangent line is the number the averages of the slopes of the secant lines approach. In this case it is 2.


2 2

:

( ) 96

16 96 6

6 2.45

The rock strikes

the water

s t

t t

t


2( ) 16

[ 6 0.1, 6]

( 6) ( 6 0.1)76.7837

6 ( 6 0.1)

s t t

Average in

s s


Area of Irregular Shapes Problem

Calculus I Chapter five 22

2For example to approximate the area under ( ) form 0 to 1:f x x

0 1area under the curve The area of the rectangle(s) overestimates the area under the curve.

1 1 50

8 2 8area under the curve

1 4 1 140

27 27 3 27area under the curve

1 1 9 1 350

32 8 64 4 64area under the curve

Calculus I Chapter five 23

If we continue with this process of dividing the interval from zero to one to more and more partitions (more rectangles), then the sum of the areas of the rectanglesbecomes closer to the exact area for every rectangle we add. If we increase the number of rectangles, hypothetically, to infinity, then the sum of the rectangles would give the exact area! Of course we cannot literally do so! But, we can do so in our Imagination using the concept of limit at infinity! Below, the number of rectangles is 10, 20,50, and 100, and the exact answer we are approaching is 1/3!Geogebra File


Limit of a Function and One-Sided Limits


Suppose the function f is defined for all x near a except possibly at a. If f (x) is arbitrarily close to a number L whenever x is sufficiently close to (but not equal to) a, then we write lim f (x) = L. x→ a


Suppose the function f is defined for all x near a but greater than a. If f (x) is arbitrarily close to L for x sufficiently close to (but strictly greater than) a, then lim f (x) = L . x→a+

Suppose the function f is defined for all x near a but less than a. If f (x) is arbitrarily close to L for x sufficiently close to (but strictly less than) a, then lim f (x) = L. x→a−


It must be true that L = M .


Geogebra file Tan(3overx)


Limit LawsLimits of Polynomial and

Rational Functions

2.3


Substitute a for x in the function!


The Limit Laws allow us to substitute 0 for h.


The statement we are trying to prove can be stated in cases as follows:For x> 0, −x ≤ x sin(1/x) ≤ x, and For x< 0, x ≤ x sin(1/x) ≤ −x.

Now for all x ≠ 0, note that −1 ≤ sin(1/x) ≤ 1 (since the range of the sine function is [−1, 1]).

For x> 0 we have −x ≤ x sin(1/x) ≤ xFor x< 0 we have −x ≥ x sin(1/x) ≥ x, which are exactly the statements we are trying to prove.


Since lim −|x| = lim |x| = 0, and since −|x| ≤ x sin(1/x) ≤ |x|, x→0 x→0

the squeeze theorem assures us that:

lim x sin(1/x) = 0 as well. x→0


c) As the speed of the ship approaches the speed of light, the observed length of the ship shrinks to 0.


How did I get the red graphgo in between of the othergraphs?

I graphed the average of the two functions!


Infinite LimitsFinding Infinite Limits Analytically


)

)

)

)

a

b

c

d


Limits at Infinity and Horizontal AsymptotesInfinite Limits at Infinity

End BehaviorEnd Behavior of sin(x) and cos(x)


4 3

2 2 2

1 sin ( ) 1x

x x x

5 0 0 5

2 2

4 3

2

1 1lim 0 lim 0

sin ( ), lim 0.

x x

x

andx x

xSo

x


Divide by the x to the largest power in the denominator:

3 3

3 6 3 32 2

4 4lim lim

1 12 (9 15 ) 2 | | (9 15 )

x x

x x

x x x xx x

GeoGebra


22 2

2 2( )

1 2 1 2(1 ) | | 1

x xf x

x xx x x x

2

2

2lim 2

1 2| | 1

2lim 2

1 2| | 1

x

x

x

xx xx

xx x

x is x when x

x is x when x


f(x) has horizontal asymptotes at y=2, y=-2


0 0

A special limit:

sin( )lim lim 1

LengOfSegmentBE

LengthOfArcBD

Geogebra file:


Show that does not exist.0

limx

x

x

0 0 0

0 0 0

lim lim lim 1 1

lim lim lim ( 1) 1

x x x

x x x

x x

x xx x

x x

Since the right- and left-hand limits are different, it follows that the limit does not exist.


4 if 4( )

8 2 if 4

x xf x

x x

4lim ( )x

f x

4 4lim ( ) lim 4 4 4 0x x

f x x

4 4lim ( ) lim (8 2 ) 8 2 4 0x x

f x x

Determine whether exists.

The right- and left-hand limits are equal. Thus, the limit exists and

4lim ( ) 0x

f x


Continuity at a pointContinuity on an intervalFunctions Involving Roots

Continuity of Trigonometric FunctionsThe Intermediate Value Theorem


2 32

2 3

( ) ( 1)

( 1) 0

1

f t t

t

t

-1 1


The Intermediate Value Theorem

The Importance of continuity condition:


A Precise Definition of LimitLimit Proofs

Infinite LimitsLimits at Infinity


:The interval around on the - that guarantees

( ) will be within of is not always symmetric.

In cases where the interval is not symmetric we pick

the smaller distance to c to be .

Notex x axis

f x L

3

1lim 5 6 2x

x x


| 2 8 2 |

| 2 6 |

| 2( 3) |

| 2( 3) |

2 | ( 3) |

| ( 3) |2

x

x

x

x

x

x

:2

Let be


10| 0 |

100 0

10 is positive

10

1

1010

x

x

x

xx

x


2

2

2

| 2 3 2 | 0.25

| 2 1| 0.25

| ( 1) | 0.25

x x

x x

x

2

2

Since both sides are positive

( 1) 0.25

( 1) 0.25

| 1| 0.25

1 0.25 1 0.25

to | 1|

1 1

= 0.25 0.5

x

x

x

x

Compare x

x

Let


2

2

2

| 2 3 2 |

| 2 1|

| ( 1) |

x x

x x

x

2

2

Since both sides are positive

( 1)

( 1)

| 1|

1 1

to | 1|

1 1

=

x

x

x

x

Compare x

x

Let


2

2

Prove the limit:

lim( 2 3) 3x

x x

2

2

| 2 3 3 |

| 2 |

| ( 2) |

x x

x x

x x

2

0Note that lim( 2 3)

is also 3!x

x x

x is near 2; let =1 (or any small positive number) so:

2 1 2 1 1 3

The largest value of x is 3 when =1, so:

| ( 2) | | 3( 2) | | ( 2) |3

Compare to | ( 2) | Let =min 1,3

x x

x x x x

x


2

0

Prove the limit:

lim( 2 3) 3x

x x

2

2

| 2 3 3 |

| 2 |

| ( 2) |

x is near 0; let =1 (or any small positive number) so:

1 1 3 2 1

The largest value of |x-2| is 5 when =1, so:

| ( 2) | | 5 | 5 | | | |5

Compare

x x

x x

x x

x x

x x x x x

to | | Let =min 1,5

x

calculus i chapter two1 to a roman a “calculus” was a pebble used in counting and in gambling....

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