calculus i chapter two1 to a roman a “calculus” was a pebble used in counting and in gambling....
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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.
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Calculus I Chapter two 2
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Calculus I Chapter two 3
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Calculus I Chapter two 4
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Calculus I Chapter two 5
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Calculus I Chapter two 6
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Calculus I Chapter two 7
Average Velocity Vs. Instantaneous Velocity
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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
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9Calculus I Chapter two
( ) 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!
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10Calculus I Chapter two
( ) 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!
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11Calculus I Chapter two
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!
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Calculus I Chapter two 12
:
( ) ( )
AverageVelocity from a to b
s b s a
b a
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Calculus I Chapter two 13
:
( ) ( )
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.
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Calculus I Chapter two 14
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].
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Calculus I Chapter two 15
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
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Calculus I Chapter two 16
Slope of the tangent line is the number the averages of the slopes of the secant lines approach. In this case it is 2.
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Calculus I Chapter two 17
2 2
:
( ) 96
16 96 6
6 2.45
The rock strikes
the water
s t
t t
t
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Calculus I Chapter two 18
2( ) 16
[ 6 0.1, 6]
( 6) ( 6 0.1)76.7837
6 ( 6 0.1)
s t t
Average in
s s
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Calculus I Chapter two 19
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Calculus I Chapter two 20
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Calculus I Chapter two 21
Area of Irregular Shapes Problem
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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
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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
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Calculus I Chapter two 24
Limit of a Function and One-Sided Limits
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Calculus I Chapter two 25
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
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Calculus I Chapter two 26
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−
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Calculus I Chapter two 27
It must be true that L = M .
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Calculus I Chapter two 28
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Calculus I Chapter two 29
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Calculus I Chapter two 32
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Calculus I Chapter two 33
Geogebra file Tan(3overx)
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Calculus I Chapter two 35
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Calculus I Chapter two 36
Limit LawsLimits of Polynomial and
Rational Functions
2.3
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Calculus I Chapter two 37
Substitute a for x in the function!
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Calculus I Chapter two 38
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Calculus I Chapter two 39
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Calculus I Chapter two 40
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Calculus I Chapter two 41
The Limit Laws allow us to substitute 0 for h.
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42Calculus I Chapter two
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43Calculus I Chapter two
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Calculus I Chapter two 44
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Calculus I Chapter two 45
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Calculus I Chapter two 46
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.
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Calculus I Chapter two 47
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
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48Calculus I Chapter two
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Calculus I Chapter two 49
c) As the speed of the ship approaches the speed of light, the observed length of the ship shrinks to 0.
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Calculus I Chapter two 50
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Calculus I Chapter two 51
How did I get the red graphgo in between of the othergraphs?
I graphed the average of the two functions!
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Calculus I Chapter two 52
Infinite LimitsFinding Infinite Limits Analytically
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Calculus I Chapter two 53
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55Calculus I Chapter two
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Calculus I Chapter two 56
)
)
)
)
a
b
c
d
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57Calculus I Chapter two
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Calculus I Chapter two 58
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Calculus I Chapter two 59
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Calculus I Chapter two 60
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Calculus I Chapter two 61
Limits at Infinity and Horizontal AsymptotesInfinite Limits at Infinity
End BehaviorEnd Behavior of sin(x) and cos(x)
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Calculus I Chapter two 62
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Calculus I Chapter two 63
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Calculus I Chapter two 64
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
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65Calculus I Chapter two
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
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66Calculus I Chapter two
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Calculus I Chapter two 67
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
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Calculus I Chapter two 68
f(x) has horizontal asymptotes at y=2, y=-2
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Calculus I Chapter two 69
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70Calculus I Chapter two
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Calculus I Chapter two 71
0 0
A special limit:
sin( )lim lim 1
LengOfSegmentBE
LengthOfArcBD
Geogebra file:
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Calculus I Chapter two 72
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.
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Calculus I Chapter two 73
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
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Calculus I Chapter two 74
Continuity at a pointContinuity on an intervalFunctions Involving Roots
Continuity of Trigonometric FunctionsThe Intermediate Value Theorem
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Calculus I Chapter two 75
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76Calculus I Chapter two
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Calculus I Chapter two 77
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78Calculus I Chapter two
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79Calculus I Chapter two
2 32
2 3
( ) ( 1)
( 1) 0
1
f t t
t
t
-1 1
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Calculus I Chapter two 80
The Intermediate Value Theorem
The Importance of continuity condition:
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Calculus I Chapter two 81
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Calculus I Chapter two 82
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Calculus I Chapter two 83
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Calculus I Chapter two 84
A Precise Definition of LimitLimit Proofs
Infinite LimitsLimits at Infinity
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Calculus I Chapter two 85
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Calculus I Chapter two 86
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Calculus I Chapter two 87
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Calculus I Chapter two 88
: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
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Calculus I Chapter two 89
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Calculus I Chapter two 90
| 2 8 2 |
| 2 6 |
| 2( 3) |
| 2( 3) |
2 | ( 3) |
| ( 3) |2
x
x
x
x
x
x
:2
Let be
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Calculus I Chapter two 91
10| 0 |
100 0
10 is positive
10
1
1010
x
x
x
xx
x
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Calculus I Chapter two 92
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
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Calculus I Chapter two 93
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
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Calculus I Chapter two 94
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
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95Calculus I Chapter two
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