Transcript Document

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.
Calculus I Chapter two
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Calculus I Chapter two
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Calculus I Chapter two
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Calculus I Chapter two
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Calculus I Chapter two
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Calculus I Chapter two
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Average Velocity Vs.
Instantaneous Velocity
Calculus I Chapter two
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In the tables below we’ll find the average rate of change of
two functions in three intervals.
f ( x)  2 x  1
x 2x 1
1
2
3
5
3
4
7
9
Calculus I Chapter two
f ( x)  x  1
2
x x 1
2
f (2)  f (1) 5  3

2
2 1
1
f (3)  f (2) 7  5

2
2 1
1
f (4)  f (3) 9  7

2
2 1
1
1
2
2
5
3
4
10
17
f (2)  f (1) 5  2

3
2 1
1
f (3)  f (2) 10  5

5
2 1
1
f (4)  f (3) 17  10

7
2 1
1
8
f ( x)  2 x  1
x 2 x  1 f (2)  f (1)  5  3  2
1
2
3
5
3
4
7
9
2 1
1
f (3)  f (2) 7  5

2
2 1
1
f (4)  f (3) 9  7

2
2 1
1
The average rate of change
DOES NOT change! In fact
for a line the average rate is
the same as instantaneous
rate, and that is the slope!
f ( x)  x  1
2
x x2  1
1
2
2
5
3
4
10
17
f (2)  f (1) 5  2

3
2 1
1
f (3)  f (2) 10  5

5
2 1
1
For this function, the
average rate of change
DOES change!
f (4)  f (3) 17  10

7
2 1
1
Calculus I Chapter two
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f ( x)  2 x  1
x 2 x  1 f (2)  f (1)  5  3  2
1
2
3
5
3
4
7
9
2 1
1
f (3)  f (2) 7  5

2
2 1
1
f (4)  f (3) 9  7

2
2 1
1
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!
f ( x)  x  1
2
x x2  1
1
2
2
5
3
4
10
17
f (2)  f (1) 5  2

3
2 1
1
f (3)  f (2) 10  5

5
2 1
1
f (4)  f (3) 17  10

7
2 1
1
But, for this function the
above question is a little
tricky! Is the rate at x=2
3 or 5 or none? This is the
question that was, finally,
answered by Calculus!
10
Calculus I Chapter two
f ( x)  x 2  1
x x2  1
1
2
2
5
3
4
10
17
f (2)  f (1) 5  2

3
2 1
1
f (3)  f (2) 10  5

5
2 1
1
f (4)  f (3) 17  10

7
2 1
1
What if we had to answer
the question without using
calculus? Well, we can
approximate the answer. But,
how? We can say the rate is
approximately 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!
Calculus I Chapter two
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Average Velocity from a to b :
s (b)  s (a )

ba
Calculus I Chapter two
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Average Velocity 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.
Calculus I Chapter two
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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].
Calculus I Chapter two
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Geogebra file: TangentLine2
f ( x)  f (a)
?
xa
when x  a
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.
Calculus I Chapter two
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Slope of the tangent line is
the number the averages of
the slopes of the secant lines
approach. In this case it is 2.
Calculus I Chapter two
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The rock strikes
the water :
s (t )  96
16t  96  t  6
2
2
t   6  2.45
Calculus I Chapter two
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s (t )  16t 2
Average in [ 6  0.1, 6]
s ( 6)  s ( 6  0.1)

 76.7837
6  ( 6  0.1)
Calculus I Chapter two
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Calculus I Chapter two
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Calculus I Chapter two
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Area of Irregular Shapes Problem
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For example to approximate the area under f ( x)  x 2 form 0 to 1:
The area of the rectangle(s) overestimates the area under the curve.
0  area under the curve  1
0  area under the curve 
1 1 5
 
8 2 8
0  area under the curve 
0  area under the curve 
Calculus I Chapter five
1
4 1 14

 
27 27 3 27
1 1 9 1 35
 
 
32 8 64 4 64
22
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 rectangles
becomes 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
Calculus I Chapter five
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Limit of a Function and
One-Sided Limits
Calculus I Chapter two
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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
Calculus I Chapter two
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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−
Calculus I Chapter two
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It must be true that L = M .
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Calculus I Chapter two
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Calculus I Chapter two
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Calculus I Chapter two
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Calculus I Chapter two
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Calculus I Chapter two
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Geogebra file Tan(3overx)
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Calculus I Chapter two
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Calculus I Chapter two
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2.3
Limit Laws
Limits of Polynomial and
Rational Functions
Calculus I Chapter two
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Substitute a for x in the function!
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Calculus I Chapter two
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Calculus I Chapter two
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The Limit Laws allow us to substitute 0 for h.
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Calculus I Chapter two
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Calculus I Chapter two
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Calculus I Chapter two
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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) ≤ x
For x< 0 we have −x ≥ x sin(1/x) ≥ x,
which are exactly the statements we are trying to prove.
Calculus I Chapter two
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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
x→0
as well.
Calculus I Chapter two
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Calculus I Chapter two
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c) As the speed of the ship approaches the speed
of light, the observed length of the ship shrinks
to 0.
Calculus I Chapter two
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Calculus I Chapter two
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How did I get the red graph
go in between of the other
graphs?
I graphed the average of
the two functions!
Calculus I Chapter two
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Infinite Limits
Finding Infinite Limits Analytically
Calculus I Chapter two
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a)  
b)  
c)  
d)  
Calculus I Chapter two
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Calculus I Chapter two
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Calculus I Chapter two
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Calculus I Chapter two
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Limits at Infinity and Horizontal Asymptotes
Infinite Limits at Infinity
End Behavior
End Behavior of sin(x) and cos(x)
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Calculus I Chapter two
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Calculus I Chapter two
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1 sin ( x ) 1

 2
2
2
x
x
x
4
3
 500  5
1
1
lim 2  0 and lim 2  0
x  x
x  x
sin 4 ( x 3 )
So, lim
 0.
2
x 
x
Calculus I Chapter two
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Divide by the x to
the largest power in
the denominator:
lim
x 
4x
3
1
2 x  x (9  15 2 )
x
3
6
 lim
x 
4x
3
1
2 x  | x | (9  15 2 )
x
3

3
GeoGebra
Calculus I Chapter two
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Calculus I Chapter two
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2x
2x
f ( x) 

1 2
1 2
2
x (1   2 ) | x | 1   2
x x
x x
2x
lim
2
x 
1 2
x is x when x  
| x | 1  2
x x
x is  x when x  
2x
lim
 2
x 
1 2
| x | 1  2
x x
Calculus I Chapter two
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f(x) has horizontal
asymptotes at y=2, y=-2
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A special limit:
lim
 0
sin( )

LengOfSegmentBE
 lim
1
  0 LengthOfArcBD
Geogebra file:
Calculus I Chapter two
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Show that lim
x 0
x
x
does not exist.
x
x
lim  lim  lim 1  1
x 0 x
x 0 x
x 0
x
x
lim  lim
 lim ( 1)  1
x 0 x
x 0
x x 0
Since the right- and left-hand limits are different, it
follows that the limit does not exist.
Calculus I Chapter two
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Determine whether
lim f ( x)
x4
exists.
 x  4 if x  4
f ( x)  
8  2 x if x  4
lim f ( x)  lim x  4  4  4  0
x4
x4
lim f ( x)  lim (8  2 x)  8  2  4  0
x4
x4
The right- and left-hand limits
are equal. Thus, the limit exists and
lim f ( x)  0
Calculus I Chapter two
x4
73
Continuity at a point
Continuity on an interval
Functions Involving Roots
Continuity of Trigonometric Functions
The Intermediate Value Theorem
Calculus I Chapter two
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Calculus I Chapter two
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Calculus I Chapter two
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78
Calculus I Chapter two
f (t )  2 (t 2  1)3
(t  1)  0
t  1
2
3
-1
1
Calculus I Chapter two
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The Intermediate Value Theorem
The Importance of
continuity condition:
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A Precise Definition of Limit
Limit Proofs
Infinite Limits
Limits at Infinity
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Calculus I Chapter two
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Note :
The interval around x on the x - axis that guarantees
f ( x) will be within  of L is not always symmetric.
In cases where the interval is not symmetric we pick
the smaller distance to c to be  .
lim x3  5 x  6  2
x 1
Calculus I Chapter two
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Calculus I Chapter two
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Let  be :  

2
| 2 x  8  2 | 
| 2 x  6 | 
| 2( x  3) | 
| 2( x  3) | 
2 | ( x  3) | 
| ( x  3) |
Calculus I Chapter two

2
90
10
 0 | 
x
10
0 
 0
x
10
is positive
x
10

x
x 1

10 
10
x
|
Calculus I Chapter two

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| x 2  2 x  3  2 | 0.25
Since both sides are positive
| x  2 x  1| 0.25
( x  1) 2  0.25
2
| ( x  1) 2 | 0.25
( x  1) 2  0.25
| x  1| 0.25
1  0.25  x  1  0.25
Compare to |x  1| 
1   x 1 
Let  = 0.25  0.5
Calculus I Chapter two
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| x  2 x  3  2 | 
2
| x  2 x  1| 
2
| ( x  1) | 
2
Since both sides are positive
( x  1) 2  
( x  1) 2  
| x  1| 
1   x 1 
Compare to |x  1| 
1   x 1 
Let  = 
Calculus I Chapter two
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Prove the limit:
lim( x  2 x  3)  3
2
x2
Note that lim( x 2  2 x  3)
x 0
is also 3!
| x 2  2 x  3  3 | 
| x 2  2 x | 
| x( x  2) | 
x is near 2; let  =1 (or any small positive number) so:
2 1  x  2 1
 1 x  3
The largest value of x is 3 when  =1, so:
| x( x  2) |   | 3( x  2) |  | ( x  2) |
 
Compare to | ( x  2) |  Let  =min 1, 
 3
Calculus I Chapter two

3
94
Prove the limit:
lim( x  2 x  3)  3
2
x 0
| x 2  2 x  3  3 | 
| x 2  2 x | 
| x( x  2) | 
x is near 0; let  =1 (or any small positive number) so:
1  x  1   3  x  2  1
The largest value of |x-2| is 5 when  =1, so:
| x( x  2) |   | 5 x |   5 | x |  | x |
 
Compare to | x |  Let  =min 1, 
 5

5
Calculus I Chapter two
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