Calculus
2.3: The Limit Theorems
Lesson 1
Warm Up
p. 43 Quick Review
Q1. Find the limit of 13x
as x approaches zero.
x
Q2. Sketch the graph of a function if 3 is the limit as x approaches 2 but f(2) is undefined.
Q3. Sketch the graph of a function that is decreasing slowly when x = -4.
Q4. Sketch the graph of a quadratic function.
Q5. Sketch the graph of y  x 3 .
Q6. Factor: x 2  100
Q7. Thirty is what percentage of 40?
Q8. What is meant by definite integral?
x3  8 x  22 x  21
Q9. Divide using synthetic division:
x 3
30
12 x
Q10. When simplified, the expression
becomes
3x10
A. 9x3
B. 9x 20
C. 4x 3
D. 4x 20
Solutions:
Q1. 13
Q2.
Q3.
y
y








x









x

















y
y








x





x














Q4.





Q5.

Q6. x 10x  10

Q7. 75%
Q8. Product of x and y, where x varies and y may vary.

3 1
Q9.
1
-8
22 - 21
3 -15
-5 7
21
0
Q10. D
x 2  5x  7
I. Properties of Limits: Direct Substitution

When direct substitution works, the limit exactly equals the value of the function. In this
section, you will learn about several cases in which direct substitution can be applied.
Direct substitution: the value of “c” can be plugged directly into the function and
evaluated.
Example 1
Evaluate the limit by direct substitution: lim x 2  1
x 3
 3
2
 1  10
This example gave us a nice real number solution. In other cases, you will get an
undefined result if you try to use direct substitution.
Example 2
Evaluate the limit by direct substitution: lim x 1 2 .
x2
Clearly, you can’t use direct substitution because that would yield
1
0
, which is undefined.
In this section, we will learn some properties and techniques for computing limits
analytically when direct substitution will not work. For all of the properties in this
section, b and c are real numbers and n is a positive integer.
Basic Limits
The three basic limits are extremely simply and you should commit them to memory right
away.
(1) lim b  b
[The limit of a constant is simply the constant].
(2) lim x  c
[This is simply direct substitution].
(3) lim x n  c n
[Again, this is nothing more than direct substitution].
x c
x c
x c
Example 3
Evaluate lim 2
x832
lim 2 = 2. Think of it like this: the function is a constant, so the graph is a horizontal
x832
line. For any value of x, the limit will be 2.
Figure 1: Graph of f(x) = 2
Example 4
Evaluate lim x
x  23
lim x = 23. This is a very easy example of direct substitution. As you know, the function
x  23
f  x   x is a line with slope 1 that passes through the origin. Sketch the situation in
Figure 2.
Figure 2: Graph of f(x) = x
Example 5
Evaluate lim x3
x 2
lim x3 = 8. This is also a simple example of direct substitution. The function f  x   x3 is
x 2
a familiar one and it is clear that the limit exists.
Figure 3: Graph of f(x) = x3
Properties of Limits
For the following five properties, lim f  x   L and lim g  x   K
xc
Property #1—Scalar multiple:
xc
lim b  f  x    bL
x c
Example 6: Scalar multiple
Let lim x3  3  5
x 2
Evaluate lim 5   x3  3 
x2
lim 5   x3  3   5  5  25
x2
This property may seem to be superfluous because, in this example, we could have just
used direct substitution without using the scalar multiple property. However, there may
be cases where a difficult limit is given in the problem statement and then you are asked
to find the limit of the difficult limit multiplied by a constant.
Another way of looking at this property is that you are allowed to take the constant out in
front of the limit:
lim b  f  x    b  lim f  x 
x c
x c
Now, use your graphing calculator to find the limit by using the VALUE feature.
Y = 25
Property #2—Sum or difference:
lim  f  x   g  x   L  K
x c
Example 7: Sum property
Evaluate the following limit by using the sum property: lim  x3  3
x 3
lim x3  lim3  27  3  30
x3
x3
Now, evaluate the limit by using direct substitution, without applying the property.
lim  x3  3  33  3  30
x 3
lim  f  x   g  x    L  K
Property #3—Product:
x c
Example 8: Product property
Given: lim f  x   14 and lim g  x   32
x2
x 2
Evaluate lim  f  x   g  x   by using the Product property
x 2
lim  f  x   g  x    14   23   83
x 2
lim
Property #4—Quotient:
x c
f  x L
 , if K  0
g  x K
Example 9: Quotient property
Given: lim f  x   14 and lim g  x   32
x2
x 2
 f  x 
Evaluate lim 
 by using the Quotient property
x2 g  x 


 f  x 
lim 

x2 g  x 


1
4
3
2

1
6
Property #5—Power:
lim  f  x   Ln
x c
n
Example 10: Power property
Given: lim f  x   14
x 2
Evaluate lim  f  x  by using the Quotient property
x 2
3
lim  f  x    14   641
x 2
3
3
Limits of Polynomial and Rational Functions
You can use direct substitution for all polynomial functions.
You can use direct substitution for all rational functions, as long as the denominator is not
zero.
Example 11: Limit of Polynomial function
Evaluate lim  4 x3  3x 2  2 x  1
x 3
lim  4 x3  3x 2  2 x  1  4  3  3  3  2  3  1  130
3
2
x 3
Example 12: Limit of Rational function
2x  3
x 3 x  5
Evaluate lim
2 x  3 2  3  3 3


x 3 x  5
 3  5 8
lim
Example 13
2x  3
x 5 x  5
Evaluate lim
2 x  3 2  5  3 13


x 5 x  5
5  5
0
lim
This means that direct substitution doesn’t work. ***You cannot conclude that the limit
doesn’t exist; all you know is that direct substitution doesn’t work.***
Limit of a Function Involving a Radical
lim n x  n c
x c
There is one important restriction on this property: If c is negative, then n must be odd.
[Recall that you can’t take an even root of a negative number—it will give you a nonreal
result].
Example 14: Limit of a Radical function
lim 3 x
x  27
lim 3 x  3
x  27
Example 15: Limit of a Composite Function
Given: f  x   2 x 2  3x  1, g  x   3 x  6
Evaluate lim g  f  x  
x 4

 

lim g  f  x    g lim f  x   g lim  2 x 2  3x  1  g  21  3
x 4
x 4
x 4
Exit Ticket
Evaluate the following limits:

1. lim  x 2 
2. lim
Solutions:
1. 4
2. undefined
x 2
x 2
x 3

Lesson 2
Warm Up
1. (T/F) If f  x   g ( x) for all real numbers other than x  0, and
lim f  x   L then lim g  x   L
x0
x0
True. These two functions agree at all but one point, so the limits are the same at that
one point.
2. If lim f  x   L, then f  c   L.
xc
False. There could be a jump at x = c.
Limit of Trigonometric Functions
You can use direct substitution for all six trigonometric functions.
lim sin x  sin c
lim cos x  cos c
lim tan x  tan c
lim cot x  cot c
lim sec x  sec c
lim csc x  csc c
x c
x c
x c
x c
x c
x c
Example 16
Evaluate lim tan x
x 
lim tan x  tan   0
x 
Example 17
Evaluate lim cos 3 x
x 
lim cos 3 x  cos 3  1
x 
II. Strategies for Finding Limits
We will commonly encounter “functions that agree at all but one point”. These functions
are identical except for the fact that one of them contains a discontinuity, such as a hole.
Figure 4: Original Polynomial function
Figure 5: Similar Rational Function
As you can see, you can have a rational function that is identical to a polynomial function
except for one point. We have a theorem that states that the limit as x approaches the one
point is equivalent for both functions:
lim f  x   lim g  x 
xc
[Functions that Agree at All but One Point]
x c
This fact is useful when dealing with rational functions that cannot be evaluated by direct
substitution.
One of the common ways of finding “functions that agree at all but one point” is to divide
out a factor from the denominator of a rational function.
Example 18
x2  x  6
Find the limit: lim
x3
x 3
Try direct substitution. Notice that you will get a zero in the denominator. In fact, you get
the indeterminate form.
Indeterminate form:
0
0
First, factor the numerator so that you get a cancellation:
x 2  x  6  x  3 x  2 

 x2
x3
 x  3
Now, you can use direct substitution:
lim x  2  5
x 3
Now, check your solution graphically. Graph the original function on your calculator:
x2  x  6
Y1 
x3
Use the TRACE feature to get close to x = -3 and take note of the y-value.
x2  x  6
 5
lim
x3
x 3
In Example 18, did we change the original function at all?
x2  x  6
Yes,
is different than x  2 at one point  x  3 . You cannot see this one
x3
point on your graphing calculator by simply looking at the graphs, but try this: use the
VALUE feature for x = -3. Now, you can tell that the point is undefined.
Another technique for dealing with the indeterminate form is called “rationalizing”.
Recall from your precalculus courses that “rationalizing the denominator” meant getting
rid of radicals in the denominators. Here, we may wish to use a similar technique to
remove radicals from the numerator in order to avoid the indeterminate form.
Example 19
Evaluate the limit:
lim
x 0
2 x  2
x
Try direct substitution: indeterminate form
Rationalize the numerator:
2 x  2

x
lim
x 0


  2  x  2 
2 x 2  x  2 x
2 x  2
2 x 
x
2 x  2


1
2 x  2
1
1
2


4
2 x  2 2 2
Example 20
Evaluate the limit:
lim
x 3
x 1  2

x 3
lim
x 3


x 1  2
x 3
   x  1  4 
x 3

x  1  2   x  3  x  1  2   x  3  x  1  2 
x 1  2
1
1

x 1  2 4
1
x 1  2
General Strategy for Finding Limits
From what we know now, here is our strategy for finding limits:
(1) Try to use Direct Substitution. [Recognize which types of limits allow it].
(2) Try to divide out a factor from the denominator.
(3) Check your solution graphically or numerically.
Multiplying by the Conjugate
Multiplying by the conjugate is a special technique for manipulating a rational function
into a useable form.
Example 21
Evaluate the following limit:
1  cos 
lim
 sin 

0
1  cos  1  cos 
1  cos 2 
sin 2 


 lim
lim
1  cos  lim
 0  sin 
 0  sin  1  cos  
 0  sin  1  cos  
 lim
 0
sin 
sin 
1
 lim
 lim
 1 12 
 1  cos    0 
 0 1  cos 
1
2
Exit Ticket Question
1. Create your own set of two functions that agree at all but one point.