Limits:
Evaluate the following limits:
(1)
SOLUTION:
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(2): Compute
SOLUTION:
(Algebraically simplify the fractions in the numerator using a common denominator.)
(Division by
is the same as multiplication by
(Factor the denominator. Recall that
)
(Divide out the factors x + 2 , the factors which are causing the indeterminate form
the limit can be computed. )
. Now
.
------------------------------------------------------------------------------------------------(3): Compute
SOLUTION
(Eliminate the square root term by multiplying by the conjugate of the numerator over itself.
Recall that
.)
(Divide out the factors x - 4 , the factors which are causing the indeterminate form
the limit can be computed. )
.
--------------------------------------------------------------------------------------------------------
(4): Compute
SOLUTION
(Recall the trigonometry identity
.)
. Now
(The numerator is the difference of squares. Factor it.)
(Divide out the factors
, the factors which are causing the indeterminate form
Now the limit can be computed. )
.
.
-------------------------------------------------------------------------------------------(5): Compute
SOLUTION
(The numerator approaches -3 and the denominator is a negative quantity which approaches 0 as
x
approaches 0 .)
(This is NOT an indeterminate form. The answer follows.)
(Thus, the limit does not exist.)
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(6) Compute
SOLUTION
(Recall that
.)
(Divide out the factors x - 1 , the factors which are causing the indeterminate form
the limit can be computed. )
. Now
.
(The numerator approaches 3 and the denominator approaches 0 as x approaches 1. However, the
quantity
in the denominator is sometimes negative and sometimes positive. Thus, the correct answer is
NEITHER
NOR
. The correct answer follows.)
The limit does not exist.
------------------------------------------------------------------------------------------------------
(7) Compute
SOLUTION
(Make the replacement
approaches 0 . )
so that
. Note that as x approaches
(Recall the well-known, but seldom-used, trigonometry identity
.)
(Recall the well-known trigonometry identity
(Recall that
)
)
,h
= 2.
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8: Compute
sin 3x
tan 3x
sin 3x
1
1
 lim cos 3x  lim



lim
x 0 sin 2 x
x 0 sin 2 x
x 0
1
cos 3x sin 2 x
3
sin 3x 
1 
2x  3
3
 lim
 lim
 lim
  (1)(1)(1) 
2  3 x0 3x  x0 cos 3x  2 x0 sin 2 x  2
2
[Note: We use the given limit lim
 0
sin 
 1 .]

-----------------------------------------------------------------------------------------------------------------3
9: Compute
lim
h 0
8h 2
1
1
 f (8) 
 . [Note: We use the definition
2
3
h
12
3 8
of the derivative f (a)  lim
h 0
f ( a  h)  f ( a )
where f ( x)  3 x
h
and a = 8.]
-----------------------------------------------------------------------------------------------------------------
cos x  1
10: Compute lim
x 
3
x 
3
 3    sin  3   
2  f 
3
2
. [Note: We use the
Definition of the derivative f ( a )  lim
xa
f ( x)  cos x And a  
3
f ( x)  f (a)
where
xa
.]
------------------------------------------------------------------------------------------------------------------------------
Graphs
Example 1 Using the given graph of g(x), find the following left- and right-hand limits.
a.
b.
c.
d.
lim g ( x)
x  0
lim g ( x)
x  0
lim g ( x)
x 1
lim g ( x)
x 1
Solution
a. This asks us to look at the graph of g(x) as x approaches 0 from the left. You can see
that the function values are getting closer and closer to -1. So,
lim g ( x)  1
x  0
b. This asks us to look at the graph of g(x) as x approaches 0 from the right. You can see
that the function values are getting closer and closer to -1. So,
lim g ( x)  1
x 0
c. This asks us to look at the graph of g(x) as x approaches 1 from the left. You can see
that the function values are getting closer and closer to -2. So,
lim g ( x)  2
x 1
d. This asks us to look at the graph of g(x) as x approaches 1 from the right. You can see
that the function values are getting closer and closer to -2. So,
lim g ( x)  2
x 1
Notice that in the solutions to parts (c) and (d) above, the function value g(1)=1 does not play a
role in determining the values of the limits. A limit is strictly the behavior of a function “near” a
point.
Example 2 Using the graph of h(x) below, find the following left- and right-hand limits.
a.
b.
lim h( x)
x  4
lim h( x)
x  4
Solution
a. Looking at the graph of h(x), as x approaches 4 from the left, you can see that the
function values keep getting more and more negative, without end. Thus, we say that
the function values approach negative infinity, written
lim h( x)  
GRAPH?
x  4
b. Looking at the graph of h(x), as x approaches 4 from the right, you can see that the
function values keep getting more and more positive without end. Thus, we say that the
function values approach positive infinity, written
lim h( x)  
GRAPH?
x  4
By considering both the left- and right-hand limits of a function as you approach a particular
value of x, you can determine whether or not the limit of the function at that point exists.
Definition of a Limit at a Point:
If lim f ( x)  L and lim f ( x)  L, then lim f ( x)  L.
x a 
x a 
x a
Therefore, if the left-hand limit does not equal the right-hand limit as x approaches a, then the
limit as x approaches a does not exist.
Example 3 Using the graph of f(x) below, find the following limits.
a.
b.
lim f ( x)
x 1
lim f ( x)
x 2
Solution
a. In previous investigations of this function, we found that
lim f ( x)  2 and lim f ( x)  2 . Therefore, by definition, since
x 1
x 1
lim f ( x)  lim f ( x)  2 , then lim f ( x)  2 .
x1
x1
x 1
It is important to notice that this limit exists even though f(1) does not exist.
b. In previous investigations of this function, we found that
lim f ( x)  1 and lim f ( x)  3 . Therefore, by definition, since
x 2
x 2
lim f ( x)  lim f ( x) , then lim f ( x) does not exist (DNE).
x 2
x 2
x 2
Example 4 Using the given graph of g(x), find lim g ( x) .
x 1
Solution In previous investigations of this function, we found that
lim g ( x)  2 and lim g ( x)  2 . Therefore, by definition, since
x 1
x 1
lim g ( x)  lim g ( x)  2 , then lim g ( x)  2 .
x 1
x 1
x 1
Example 5 Using the given graph of h(x), find lim h( x) .
x4
Solution: In previous investigations of this function, we found that
lim h( x)   And lim h( x)   . Therefore, by definition, since
x  4
x 4
lim h( x)  lim h( x) , then lim h( x ) DNE.
x 4
x 4
x4
So far we have been focusing on what is happening with functions at particular values of x by
looking at what is happening to the function values corresponding to values very near to the x
value. Let’s now explore what happens to the function values when we allow x to approach
positive and negative infinity.
Figure 3.4
Figure 3.4a
Figure 3.4b
Figure 3.4c
In Figure 3.4a, you can see that if you move to the right on the graph and allow x to continually
become larger (approach infinity), the function values also become larger and larger. If you
move to the left on the graph and allow x to become more and more negative (approach
negative infinity), you can see that the function values are again becoming larger and larger.
Thus, we have
lim k ( x)  
x 
and
lim k ( x)  
x 
In Figure 3.4b, you can see that if you allow x to approach infinity, the function values go
towards negative infinity. If you allow x to approach negative infinity, you can see that the
function values go towards positive infinity. Thus, we have
lim m( x)  
x 
and
lim m( x)  
x 
In Figure 3.4c, you can see that if you allow x to approach either positive or negative infinity,
the function values approach zero. Thus, we have
lim p ( x)  0
x 
and
lim p( x)  0
x 
When a function approaches a numerical value, say L, as x   or as x   , we say that the
function has a horizontal asymptote at y = L. Thus, we have just found that p(x) has a horizontal
asymptote at y=0, while k(x) and m(x) have no horizontal asymptotes.
(Note: You can only approach positive infinity from the left and you can only approach negative
infinity from the right so there is no discussion of left- and right-hand limits at infinity.)
Example 6 Using the graph of f(x) below, find the following limits.
a.
b.
c.
lim f ( x)
x 5
lim f ( x )
x 
lim f ( x)
x 0
d. lim f ( x)
x 
Solution
a. We need to find and compare the left- and right-hand limits of f(x) at x= -5. As x
approaches -5 from the left, f(x) approaches 1 and as x approaches -5 from the right, f(x)
also approaches 1. Therefore,
lim f ( x)  1
x 5
b. As x   , the function values get more and more positive without end. Therefore,
lim f ( x)  
x 
c. As x approaches zero from the left, f(x) approaches 5. Therefore,
lim f ( x)  5
x 0
d. As x   , the function values get closer and closer to zero. Therefore
lim f ( x)  0
x 
-------------------------------------------------------------------Table:
 4, x  1
Example 7 Using tables, find the following limits given that f ( x)   2
x , x  1
a. lim f ( x )
x2
b. lim f ( x )
x 1
Solution
a. We need to construct a table with x-values approaching 2 from both sides. Since all of
these x-values are in the domain of x >1, we will use the part of the function defined by
x 2 to determine the function values in our table.
Limit from the left


Limit from the right


x
1.99
1.999
1.9999
2
2.0001
2.001
2.01
f(x)
3.96
3.996
3.9996
?
4.0004
4.004
4.04
Approaching x=2 from both the left and the right sides shows that the function
values
are approaching 4. Thus, lim f ( x)  4 .
x2
b. We need to construct a table with x-values approaching 1 from both sides. All x-values
approaching x = 1 from the left are in the domain x < 1, so we will be using the part of
the function defined by 4 when finding these function values. All x-values approaching x
= 1 from the right are in the domain x > 1, so we will use x 2 to find these function
values in our table.
Limit from the left


x
f(x)
Limit from the right


0.99
0.999
0.9999
1
1.0001
1.001
1.01
4
4
4
?
1.0002
1.002
1.0201
As x 1 from the left, f(x) seems to be approaching 4, while as x 1 from the right,
f(x) seems to be approaching 1. Since these are not equal, by definition,
lim f ( x ) does not exist.
x 1
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Exercise 3.5 Fill in the given table and use it to find lim f ( x ) .
x2
x
f ( x) 
1.99
1.999
1.9999
2
x2
2.0001
2.001
2.01
?
x2
----------------------------------------------------------------------------------------------------------------------------Exercise 3.6 Fill in the given table and use it to find lim g ( x) .
x 0
x
-0.01
-0.001
-0.0001
 x  1, x  0

g ( x)  3,
x0
2
 x  1, x  0

0
0.0001
?
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Vertical asymptotes and infinite limits:
0.001
0.01
2)
3)
4)
---------------------------------------------------------------------Continuity:
1) Types of discontinuities:
2)
3)
4)
-------------------------------------------------------------------------
Determining continuity from a graph or rule for a
function:
1)
2)
3)
4)
Intermediate Value Theorem: numerical. Not found.
DAY2========================================
Compute derivatives of functions: power, exponential,
logarithmic, trigonometric, inverse trig
Apply Product Rule, Quotient Rule, Chain Rule, etc.
1)
Solution:
2)
Apply Product Rule, Quotient Rule, Chain Rule, etc:
1)
Understand the first and second derivative graphically:
Approximate derivative from graph or tables:
2)
Solution:
Interpretation of the derivative as a rate of change (limit
of an average rate of change):
2)
3)
Solution:
Relationship between differentiability and continuity:
Solution: