Limits and Rates of Change

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Calculus I Mrs. Farber
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))
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Area
2. 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
y f(x)
x
a
b
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From BC (before calculus)
We can calculate the slope of a line given two
points
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.
.
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Slope of Secant line PQ
P(0.5, 0.5)
Point Q
x f(x)
.1 .99
.2 .98
.3 .92
.4 .76
Let x values get closer and closer to .5.
Determine f(x) values.
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Slope of Secant line PQ
As Q gets closer to P, the Slope of the secant
line PQ Gets closer and closer to the slope Of
the line tangent to the Curve at P.
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Slope of a curve at a point
Figure 1.4 The tangent line at point P has the
same steepness (slope) that the curve has at P.
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)
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Slope formula
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.
.
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Slope
Find the slope of y x2 at the point (1,1) Find
the equation of the tangent line.
A
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Find slope of tangent line on f(x) x2 at the
point (1,1)
Approaching x 1 from the right
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
Slope appears to be getting close to 2.
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Find slope of tangent line on f(x) x2 at the
point (1,1)
Approaching x 1 from the left
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
Slope appears to be getting close to 2.
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Write the equation of tangent line

• As 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

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Average rate of change (from bc)
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.
Find the average velocity if f (t) 2 cost
on 0, ?
1. Calculate the function value (position) at
each endpoint of the interval
f(?) 2 cos (?) 2 1 1
f(0) 2 cos (0) 2 1 3
2. Use the slope formula
The average velocity on on 0, ? is
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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.
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2.2
To understand the instantaneous rate of change
(slope) problem and the area problem, you will
need to learn about limits
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Limits
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?

• We write this as

• The answer can be found graphically, numerically
  and analytically.

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Graphical Analysis
f (x)
x
What happens to f(x) as x gets closer to 2?
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Numerical Analysis
Use one sided limits
Start to the left of 2 and choose x values
getting closer and closer (but not equal) to 2
x
1.5
1.9
1.99
1.999
1.9999
f (x)
9.25
11.41
11.941
11.994001
11.99940001
Could x get closer to 2? Does f(x) appear to get
closer to a fixed number?
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Numerical Analysis
Start to the right of 2 and choose x values
getting closer and closer (but not equal ) to 2
x
2.5
2.1
2.01
2.001
2.0001
f (x)
15.3
12.61
12.0601
12.006001
12.00060001
If the limit exists, f(x) must approach the same
value from both directions. Does the limit exist?
Guess what it is.
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Limits that do not exist
Figure 1.8 The functions in Example 7.
In order for a limit to exist, the function must
approach the same value From the left and from
the right.
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Infinite Limits
What happens to the function value as x gets
closer and closer to 3 from the right?
x 3.5 3.1 3.01 3.001 3.0001 3.00001 3.000001
y 3 11 101 1001 10001 100001 1000001
The function increases without bound so we say
There is a vertical asymptote at x 3.
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The line xa is a Vertical Asymptote if at least
one is true.
Identify any vertical asymptotes
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Graph of f(x)
True or false

1. x 2 is in the domain of f

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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
Example
then
Evaluate by direct substitution 2-5 -3
As x gets closer and closer and closer to 2, the
function value gets closer and closer to -3.
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Analytic

Using direct substitution,
As x gets closer and closer to 2 (but not equal
to 2) f(x) gets closer and closer to 12
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Compute some limits
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Basic Limits
If b and c are real numbers and is n a positive
integer
Guess an answer and click to check.
Guess an answer and click to check.
Guess an answer and click to check.
-2
Ex
5
Ex
9
Ex
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Properties of Limits
Multiplication by a constant b
Limit of a sum or difference
Limit of a product
Limit of a power
Limit of a quotient when denominator is not 0.
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Using Properties of Limits
Properties allow evaluation of limits by direct
substitution for many functions.
Ex.
As x gets closer and closer to 3, the function
value gets closer and closer to 9.
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Analytic Techniques

• Direct substitution
• First 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.

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Analytic Techniques

• Rewrite algebraically if direct substitution
  produces an indeterminate form such as 0/0
• Factor and reduce
• Rationalize a numerator or denominator
• Simplify 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.
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Find the indicated limit
direct substitution fails
Rewrite and cancel
- 5
now use direct sub.
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Find the indicated limit
direct substitution fails
Rewrite and cancel
now use direct sub.
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Find the indicated limit
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calculate one sided limits
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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
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Determine the limit on y sin ?/? as ?
approaches 0.
Figure 1.24 The graph of f (?) (sin ?)/?.
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A one-sided limit
Figure 1.37 The graph of y e1/x for x lt 0
shows limx?0 e1/x 0. (Example 11)
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Limits that are infinite (y increases without
bound)
An infinite limit will exist as x approaches a
finite value when direct substitution produces
If an infinite limit occurs at x c we have a
vertical asymptote with the equation x c.
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2.5 Continuity in (a) at x 0 but not in other
graphs.
Figure 1.50 The function in (a) is continuous
at x 0 the functions in (b) through ( f ) are
not.
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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.

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Find the reasons for discontinuity in b, c, d, e
and f.
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Composite Functions
Figure 1.53 Composites of continuous functions
are continuous.
If two functions are continuous at x c then
their composition will be continuous.
is continuous for all reals.
Example
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Exploring Continuity
Are there values of c and m that make the
function continuous At x 1? Find c and m or
tell why they do not exist.
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Exploring Continuity
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2.6 Slope of secant line and slope of tangent line
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s(t) 8(t3 6 t2 12t)
Position of a car at t hours.
1. Draw a graph.
2. Does the car ever stop?
3. What is the average velocity for the following
intervals a. 0, 2, b. .5, 1.5 c. .9,1.1
4. Estimate the instantaneous velocity at t 1
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s(t) 8(t3 6 t2 12t)
2. Appears to stop at t 2. (Velocity 0)
3. What is the average velocity for 0, 2, .5,
1.5 .9,1.1
t s(t)
0 0
2 62
.5 37
1.5 63
.9 53.352
1.1 58.168
a) 31 mph b) 26 mph c) 24.08 mph
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Find an equation of the tangent line to y 2x3
4 at the point P(2, 12)
So, m 24. Use the point slope form to write the
equation
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Slope of the tangent line at x a
Figure 1.62 The tangent slope is
f (x0 h) f (x0) h
lim
h?0
Q(a h, f (a h))
f(ah) f(a)
P(a, f(a))
a h
a
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Other form for Slope of secant line of tangent
line
Let h x - a
Then x a h
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Find an equation of the tangent line at (3, ½) to
At a 3, m - 1/8
Using the point-slope formula