Rate of ChangeAnd Limits
What is Calculus?
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Two Basic Problems of Calculus
1. Find the slope of the curve y = f (x) at the point (x, f (x))
(x, f(x))
(x, f(x))
(x, f(x))
Area2. Find the area of the region bounded above by the curve y = f(x), below by the x-axis and by the vertical lines x = a and x = b
a b
y = f(x)
x
From BC (before calculus)We can calculate the slope of a line given two points
2 1
2 1
change in y y y yslope
change in x x x x
Calculate the slope of the line between the given point P (.5, .5) and another point on the curve, say Q(.1, .99). The line is called a secant line.
.
.99 .5 .491.225
.1 .5 .4slope
Slope of Secant line PQ
x f(x)
.1 .99
.2 .98
.3 .92
.4 .76
Point Q
P(0.5, 0.5)
Let x values get closer and closer to .5. Determine f(x) values.
Slope of Secant line PQ
As Q gets closer to P, theSlope of the secant line PQGets closer and closer to the slope Of the line tangent to the Curve at P.
Figure 1.4: The tangent line at point P has the same steepness (slope) that the curve has at P.Slope of a curve at a point
The slope of the curve at a point P is defined to be the slope of the line that is tangent to the curve at point P.In the figure the point is P(0.5, 0.5)
2 1
2 1
change in y y y yslope
change in x x x x
In calculus we learn how to calculate the slope at a given point P. The strategy is to take use secant lines with a second point Q. and find the slope of the secant line.
Continue by choosing second points Q that are closer and closer to the given point P and see if the difference quotient gets closer to some fixed value.
.
Slope formula
A
Find the slope of y = x2 at the point (1,1) Find the equation of the tangent line.
left thefrom approaches PQ
right thefrom approaches PQ
Slope
Find slope of tangent line on f(x) =x2 at the point (1,1)
x f(x) Slope of secant between (1,1) and (x, f(x))
2 4 3
1.5 2.25 2.5
1.1 1.21 2.1
1.01 1.021 2.01
1.001 1.002001 2.001
Approaching x = 1 from the right
Slope appears to be getting close to 2.
Find slope of tangent line on f(x) =x2 at the point (1,1)
x f(x) Slope of secant between (1,1) and (x, f(x))
0 0 1
.5 .25 1.5
.9 .81 1.9
.99 .9801 1.99
.999 .998001 1.999
Approaching x = 1 from the left
Slope appears to be getting close to 2.
Write the equation of tangent lineAs the x value of the second point
gets closer and closer to 1, the slope gets closer and closer to 2. We say the limit of the slopes of the secant is 2. This is the slope of the tangent line.
To write the equation of the tangent line use the point-slope formula
1 1( )
1 2( 1)
2 1
y y m x x
y x
y x
Average rate of change (from bc)2 1
2 1
Average rate of changechange in y y y y
change in x x x x
Find the average velocity if f (t) = 2 + cost on [0, ]
f() = 2 + cos () = 2 – 1 = 1f(0) = 2 + cos (0) = 2 + 1 = 3
2 1
2 1
1 3 2
0
y y y
t t t
1. Calculate the function value (position) at each endpoint of the interval
The average velocity on on [0, ] is2
.6366
If f(t) represents the position of an object as a function of time, then the rate of change is the velocity of the object.
2. Use the slope formula
Instantaneous rate of change
To calculate the instantaneous rate of change of we could not use the slope formula since we do not have two points.
To approximate instantaneous calculate the average rates of change in shorter and shorter intervals to approximate the instantaneous rate of change.
To understand the instantaneous rate of
change (slope) problem and the area problem, you will need to learn
about limits
2.2
Limits
We write this as:
The answer can be found graphically, numerically and analytically.
2
8lim
3
2
x
xx
2
8)(
3
x
xxf
What happens to the value of f (x) when the value of x gets closer and closer and closer (but not necessarily equal) to 2?
Graphical Analysis
2
8lim
3
2
x
xx
5 4 3 2 1 0 1 2 3 4 542
2468
101214161820
f (x)
x
What happens to f(x) as x gets closer
to 2?
Numerical Analysis2
8lim
3
2
x
xx
2
8lim
3
2
x
xx
Start to the left of 2 and choose x values getting closer and closer (but not equal) to 2
x
f (x)
1.5
9.25
1.9
11.41
1.99
11.941
1.999
11.994001
1.9999
11.99940001
Use one sided limits
Could x get closer to 2? Does f(x) appear to get closer to a fixed number?
Numerical Analysis2
8lim
3
2
x
xx
2
8lim
3
2
x
xx
If the limit exists, f(x) must approach the same value from both directions. Does the limit exist? Guess what it is.
Start to the right of 2 and choose x values getting closer and closer (but not equal ) to 2
x
f (x)
2.5
15.3
2.1
12.61
2.01
12.0601
2.001
12.006001
2.0001
12.00060001
Figure 1.8: The functions in Example 7.
Limits that do not exist
In order for a limit to exist, the function must approach the same valueFrom the left and from the right.
Infinite Limits
3
2lim
3x
x
x
What happens to the function value as x gets closer and closer to 3 from the right?
The function increases without bound so we say
3
2lim
3x
x
x
There is a vertical asymptote at x = 3.
x 3.5
3.1 3.01 3.001
3.0001
3.00001
3.000001
y 3 11 101 1001 10001 100001
1000001
4939291991
1121314151
The line x=a is a Vertical Asymptote if at least one is true.
lim ( )x a
f x
lim ( )x a
f x
lim ( )x a
f x
lim ( )x a
f x
Identify any vertical asymptotes:
2( )
5
xf x
x
2
2( )
5 6
xf x
x x
2
2( )
5 6
xf x
x x
2
2( )
6
xf x
x
x 7 6.999 1.80t 2.2 2.205 7Graph of f(x)
(a)x = 2 is in the domain of f
True or false
2limx
exists
(b)
2 2lim ( ) lim ( )
x xf x f x
(c)
2.3 Functions That Agree at All But One Point
If f(x) = g(x) for all x in an open interval except x = c then:
)(lim)(lim xgxf cxcx
252
1072
xifx
x
xx
)5(lim2
107lim 2
2
2
x
x
xxxx
Evaluate by direct substitution 2-5 = -3
then
Example
As x gets closer and closer and closer to 2, the function value gets closer and closer to -3.
Analytic
2
)42)(2(lim
2
2
x
xxxx
2
8lim
3
2
x
xx
Using direct substitution,
124)2(222
As x gets closer and closer to 2 (but not equal to 2) f(x) gets closer and closer to 12
=
)42(lim 22 xxx=
Compute some limits
2
2
4lim
2x
x
x
3
0
8lim
2x
x
x
3
3
8lim
2x
x
x
3
2
8lim
2x
x
x
Basic LimitsIf b and c are real numbers and is n a positive integer
1. bbcx lim
Ex: 2lim 7 x
2. cxcx lim
Ex: xx 5lim
3. nncx cx lim
Ex: 2
3lim xx
= -2
= 5
= 9
Guess an
answer and
click to check.
Guess an
answer and
click to check.
Guess an
answer and
click to check.
)(lim)(lim)()(lim xgxfxgxf axaxax
)(lim*)(lim)(*)(lim xgxfxgxf axaxax
)(lim)(lim xfbxbf axax
naxn
ax xfxf )(lim)(lim
0)(lim,)(lim
)(lim
)(
)(lim
xg
xg
xf
xg
xfax
ax
axax
Multiplication by a constant b
Limit of a sum or difference
Limit of a product
Limit of a power
Limit of a quotientwhen denominatoris not 0.
Properties of Limits
Properties allow evaluation of limits by direct substitution for many functions.
Ex.:6lim
)2(3lim
6
)2(3lim
3
23
2
3
x
xx
x
xx
x
xx
)6(lim
)2(lim*3lim
3
32
3
x
xx
x
xx
)6(lim
)2(lim*)(lim3
3
32
3
x
xx
x
xx
93
)1)(9(3
)63(
)23(*)3(3 2
As x gets closer and closer to 3, the function value gets closer and closer to 9.
Using Properties of Limits
Analytic Techniques
Direct substitutionFirst substitute the value of x being approached into the function f(x). If this is a real number then the limit is that number.If the function is piecewise defined, you must perform the substitution from both sides of x. The limit exists if both sides yield the same value. If different values are produced, we say the limit does not exist.
Analytic Techniques
Rewrite algebraically if direct substitution produces an indeterminate form such as 0/0
Factor and reduceRationalize a numerator or
denominatorSimplify a complex fraction
When you rewrite you are often producing another function that agrees with the original in all but one point. When this happens the limits at that point are equal.
Find the indicated limit2
3
6lim
3x
x x
x
3lim ( 2)
xx
3
( 3)( 2)lim
3x
x x
x
= - 5
direct substitution fails
Rewrite and cancel
now use direct sub.
0
0
Find the indicated limit
0
1 1limx
x
x
direct substitution fails
Rewrite and cancel
now use direct sub.
0
0
0 0
1 1 1 1lim * lim
1 1 [ 1 1]x x
x x x
x x x x
0
1 1lim
21 1x x
Find the indicated limit2 1, 2
( )5 3, 2
x xf x
x x
calculate one sided limits
7
5
2lim ( )x
f x
Since the one-sided limits are not equal, we say the limit does not exist. There will be a jump in the graph at x =2
2lim ( ) 5
xf x
2lim ( ) 7
xf x
Figure 1.24: The graph of f () = (sin )/.Determine the limit on y = sin θ/θ as θ approaches 0.
Although the function is not defined at θ =0, the limit as θ 0 is 1.
Figure 1.37: The graph of y = e1/x for x < 0 shows limx0
– e1/x = 0. (Example 11)A one-sided limit
0 0limx
Limits that are infinite (y increases without bound)
41
lim4x x
21
lim2x x
31
lim3x x
An infinite limit will exist as x approaches a finite value when direct substitution produces
0
not zero
If an infinite limit occurs at x = c we have a vertical asymptote with the equation x = c.
Figure 1.50: The function in (a) is continuous at x = 0; the functions in (b) through ( f ) are not.2.5 Continuity in (a) at x = 0 but not in other graphs.
Conditions for continuity
A function y = f(x) is continuous at x = c if and only if:• The function is defined at x = c• The limit as x approaches c exists• The value of the function and the value of the limit are equal.
( ) lim ( )x c
f c f x
Find the reasons for discontinuity in b, c, d, e and f.
Figure 1.53: Composites of continuous functions are continuous.
Composite Functions
Example:2( ) 4f x x is continuous for all reals.
If two functions are continuous at x = c then their composition will be continuous.
Exploring Continuity
2
3
1
4 1
1
cx if x
if x
x mx if x
Are there values of c and m that make the function continuousAt x = 1? Find c and m or tell why they do not exist.
Exploring Continuity2
3
1
4 1
1
cx if x
if x
x mx if x
(1) 4f
2
1lim ( ) (1)x
f x c c
3
1lim ( ) (1) (1) 1x
f x m m
4c
1 4
5
m
m
2.6 Slope of secant line and slope of tangent line
sec( ) ( )y f x f a
mx x a
tan( ) ( )
limx af x f a
mx a
s(t) = 8(t3 – 6 t2 +12t)
1. Draw a graph.
3. What is the average velocity for the following intervals a. [0, 2], b. [.5, 1.5] c. [.9,1.1]
2. Does the car ever stop?
4. Estimate the instantaneous velocity at t = 1
Position of a car at t hours.
t s0 01 562 643 72
s(t) = 8(t3 – 6 t2 +12t)
3. What is the average velocity for[0, 2], [.5, 1.5][.9,1.1]
0 0.5 1 1.5 2 2.5 3
1020304050607080
2. Appears to stop at t =2. (Velocity= 0)
t s(t)
0 0
2 62
.5 37
1.5 63
.9 53.352
1.1 58.168
a) 31 mphb) 26 mphc) 24.08 mph
Find an equation of the tangent line to y = 2x3 – 4 at the point P(2, 12)
tan( ) ( )
limx af x f a
mx a
3 3
2 2(2 4) 12 2 16
lim lim2 2x x
x x
x x
2
22( 2)( 2 4)
lim2x
x x x
x
22lim 2( 2 4) 24x x x
12 24( 2)y x 24 36y x
So, m = 24. Use the point slope form to write the equation
Figure 1.62: The tangent slope is
f (x0 + h) – f (x0)hh0
lim
Slope of the tangent line at x= a
f(a+h) – f(a)
a a + h
P(a, f(a))
Q(a + h, f (a + h))
Other form for Slope of secant line of tangent line
sec( ) ( )y f a h f a
mx h
tan 0( ) ( )
limhf a h f a
mh
Let h = x - a Then x = a + h
tan 0( ) ( )
limhf a h f a
mh
Find an equation of the tangent line at (3, ½) to
2
1y
x
0
2 21 1limh
a h ah
0
2( 1) 2( 1)lim
( 1)( 1)ha a h
h a a h
0 02 2 2 2 2 2
lim lim( 1)( 1) ( 1)( 1)h h
a a h h
h a a h h a a h
0 2
2 2lim
( 1)( 1) ( 1)h a a h a
At a = 3, m = - 1/8
Using the point-slope formula:
1 1( 3)
2 8y x
1 7
8 8y x
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