Make a Stick Activity
• Use markers/ colored pencils/ or whatever
you like to write your name on a stick and
decorate it to make it your own.
• When you’re done, bring your stick up to the
front table and put it in the box for your class.
Welcome To Calculus BC
Expectations
•
•
•
•
•
•
•
Effort
Ask Questions – All are valid
RESPECT- Support and help each other
Return books to back shelf
Bring: notebook, pencil, red pen, calculator
Backpacks & purses under desk or in back of room
Cell phones on silent and not on your person
My teacherpage: http://www.husd.org//Domain/1454
AP Expectations
• This class is an AP course. Everyone
takes the AP test in May.
• You will need to spend a significant
amount of time outside of class on
homework, AP review, and preparation for
the AP test.
1
Limits and Their Properties
Copyright © Cengage Learning. All rights reserved.
1.2-1.4A
BC Day 1
Limits Review
Handout: Limits and their Properties
Notes
Copyright © Cengage Learning. All rights reserved.
Objectives
 Estimate limits numerically, graphically and
algebraically.
 Learn different ways that a limit can fail to
exist.
 Special Trig Limits
 Define Continuity.
Formal definition of a Limit:
If f(x) becomes arbitrarily close to a single number
L as x approaches c from either side, the limit of
f(x), as x approaches c, is L.
This limit is written as
“The limit of
lim f  x   L
x c
f of x as x approaches c is L.”
Consider the graph of f(θ) = sin(θ)/θ
• Let’s fill in the following table
θ
sin(θ)/θ
0.5
0.959
0.4
0.974
0.3
0.985
0.2
0.993
0.1
0.05
0.998 0.9995
• We can say that the limit of f(θ) approaches 1 as
θ approaches 0 from the right
• We write this as
sin 
lim
 0

1
• We can construct a similar table to show what
happens as θ approaches 0 from the left
θ
sin(θ)/θ
-0.5
0.959
-0.4
0.974
-0.3
0.985
-0.2
0.993
-0.1
-0.05
0.998 0.9995
• So we get lim
 0
sin 

1
sin 
sin 
 lim
1
• Now since we have lim

0
 0


we say that the limit exists and we write
sin 
lim
1
 0

An Introduction to Limits
Ex: Find the following limit:
x 1
lim
x 1 x  1
3

An Introduction to Limits
Start by sketching a graph of the function
For all values other than x = 1, you can use standard
curve-sketching techniques.
However, at x = 1, it is not clear what to expect.
We can find this limit numerically:
An Introduction to Limits
To get an idea of the behavior of the graph of f near x = 1,
you can use two sets of x-values–one set that approaches
1 from the left and one set that approaches 1 from the right,
as shown in the table.
An Introduction to Limits
The graph of f is a parabola that
has a gap at the point (1, 3), as
shown in the Figure 1.5.
Although x can not equal 1, you
can move arbitrarily close to 1,
and as a result f(x) moves
arbitrarily close to 3.
Using limit notation, you can write
Figure 1.5
This is read as “the limit of f(x) as x approaches 1 is 3.”
This discussion leads to an informal
definition of a limit:
A limit is the value (meaning y value) a
function approaches as x approaches
a particular value from the left and from
the right.
Properties of Limits:
Limits can be added, subtracted, multiplied, multiplied
by a constant, divided, and raised to a power.
For a limit to exist, the function must approach the
same value from both sides.
lim f  x   L
x c
One-sided limits approach from either the left or right
side only.
lim f  x   L
x c 
lim f  x   L
x c 

y  f  x
2
1
1
2
3
lim f  x  
x
4
Limits That Fail to Exist - 3 Reasons
Discuss the existence of the limit:
1
lim 2 
x 0 x
Solution: Using a graphical
representation, you can see
that x does not approach any
number. Therefore, the limit
does not exist.
lim  
x 0
lim  
x 0
lim DNE
x 0
Properties of Limits
Properties of Limits
Properties of Limits
Compute the following limits
1
lim
x 2 x  1
lim tan x
x3
lim
x 3 x  1
lim x cos x
x

4
x 
sin 2 x
lim
x 
x
x 1
lim 2
x 1 x  x  2
• Let’s take a look at the last one
x 1
lim 2
x 1 x  x  2
• What happened when we plugged in 1 for x?
0
• When we get
we have what’s called an
0
indeterminate form
• Let’s see how we can solve it
x 1
• Let’s look at the graph of lim 2
x 1 x  x  2
Is the function continuous at x = 1?
Strategies for Finding Limits?
You Try:
Find the limit:
5
You Try:
x  27
Evaluate lim
x 3 x  3
3
27
Example – Rationalizing Technique
Find the limit:
Solution:
By direct substitution, you obtain the indeterminate form
0/0.
Solution
cont’d
In this case, you can rewrite the fraction by rationalizing the
numerator.
Solution
Now, using Theorem 1.7, you can evaluate the limit
as shown.
cont’d
Solution
cont’d
A table or a graph can reinforce your conclusion that the
limit is . (See Figure 1.20.)
Figure 1.20
Solution
cont’d
You Try:
Evaluate lim
x 0
x4 2
x
1
4
Example:
Evaluate lim f  x 
x 0
 x  1, x  0
f  x   2
 x  1, x  0
lim f  x   1
x 0 
lim f  x   1
x 0 
SPECIAL TRIG LIMITS
You must know these for the AP test!
Example – A Limit Involving a Trigonometric Function
Find the limit:
Solution:
Direct substitution yields the indeterminate form 0/0.
To solve this problem, you can write tan x as (sin x)/(cos x) and obtain
Solution
Now, because
you can obtain
cont’d
Solution
cont’d
Figure 1.23
AP example
sin x
lim
1
x 0
x
• Find the following:
sin 4 x
Lim
x 0
x
4
Continuity at a Point and on an Open Interval
Continuity at a Point and on an Open Interval
Figure 1.25 identifies three values of x at which the graph of
f is not continuous. At all other points in the interval (a, b),
the graph of f is uninterrupted and continuous.
Figure 1.25
Most of the techniques of calculus require that functions be
continuous. A function is continuous if you can draw it in one motion
without picking up your pencil.
A function is continuous at a point if the limit is the same
as the value of the function.
This function has discontinuities at x=1 and
x=2.
2
1
1
2
3
4
It is continuous at x=0 and x=4, because
the one-sided limits match the value of the
function

Continuity at a Point and on an Open Interval
Consider an open interval I that contains a real number c.
If a function f is defined on I (except possibly at c), and f is
not continuous at c, then f is said to have a discontinuity
at c.
Discontinuities fall into two categories: removable and
nonremovable.
A discontinuity at c is called removable if f can be made
continuous by appropriately defining (or redefining f(c)).
Continuity at a Point and on an Open Interval
For instance, the functions shown in Figures 1.26(a) and (c)
have removable discontinuities at c and the function shown
in Figure 1.26(b) has a nonremovable discontinuity at c.
Figure 1.26
Example 1 – Continuity of a Function
Discuss the continuity of each function.
Example 1(a) – Solution
The domain of f is all nonzero real numbers.
From Theorem 1.3, you can conclude that f is continuous at
every x-value in its domain.
At x = 0, f has a non removable
discontinuity, as shown in
Figure 1.27(a).
In other words, there is no way to
define f(0) so as to make the function
continuous at x = 0.
Figure 1.27(a)
Example 1(b) – Solution
cont’d
The domain of g is all real numbers except x = 1.
From Theorem 1.3, you can conclude that g is continuous at
every x-value in its domain.
At x = 1, the function has a removable
discontinuity, as shown in
Figure 1.27(b).
If g(1) is defined as 2, the
“newly defined” function is
continuous for all real numbers.
Figure 1.27(b)
Example 1(c) – Solution
The domain of h is all real numbers. The function h is
continuous on
and
, and, because
h is continuous on the entire real line, as shown in
Figure 1.27(c).
Figure 1.27(c)
cont’d
,
Example 1(d) – Solution
The domain of y is all real numbers. From Theorem 1.6,
you can conclude that the function is continuous on its
entire domain,
, as shown in Figure 1.27(d).
Figure 1.27(d)
cont’d
Removing a discontinuity:
x3  1
f  x  2
x 1
has a discontinuity at
x 1
Write an extended function that is continuous at
 x  1  x 2  x  1 1  1  1
x3  1
 lim

lim 2
x 1
x 1 x  1
 x  1 x  1
2
 x3  1
 2 , x  1
f  x    x 1
 3 , x 1
 2

x 1
3
2
Note: There is another discontinuity
at x  1 that can not be removed.
Removing a discontinuity:
5
4
3
2
1
-5 -4 -3 -2 -1 0
-1
1
2
3
4
5
-2
-3
-4
-5
 x3  1
 2 , x  1
f  x    x 1
 3 , x 1
 2

Group Work:
Sketch the graph of f. Identify the
values of c for which lim f  x  exists.
x c
lim f  x  exists for all values except where c  4.
x c
Homework
Limits and Continuous Functions WS
Get Books! This ppt is on my teacher-page.