Download Lesson 4

Lesson 4 One-Sided Limits
Example Find xlim
4
x4
Based on the discussion and work in Lesson 3, most students say that
lim x  4  0 .
x4
However, xlim
4
Notation:
x  4 = DNE (Does Not Exist).
x  a  means x approaches a from the right-hand side of a .
Hence, x > a.
x  a  means x approaches a from the left-hand side of a .
Hence, x < a.
Left-hand side of a
x  a
Right-hand side of a
x  a

a
Definition (Right-Hand Limit) Let f be a function defined on the interval (a, a +
r), where r > 0. (Note: the function f is defined on the right-hand side of x = a.) If

as x  a , f ( x )  M , then we write lim  f ( x )  M .
xa
Definition (Left-Hand Limit) Let f be a function defined on the interval (a  r,
a), where r > 0. (Note: the function f is defined on the left-hand side of x = a.) If

as x  a , f ( x )  N , then we write lim  f ( x )  N .
xa
Examples Find the following limits, if they exist.
1.
lim ( 4 x  5 )
x  2
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
We need to be able to approach 2 from the right-hand side. Thus, the
function y  4 x  5 needs to be defined on some open interval of the form
( 2 , 2  r ) , where r > 0, so that we can approach 2 from the right-hand side.
Since the function y  4 x  5 is a polynomial, then its domain is the set of
all real numbers. Thus, we can approach 2 from the right-hand side. Since as
x  2  , 4 x  5  3 , then lim  ( 4 x  5 )  8  5  3 .
x2
Answer: 3
Animation of this limit.
NOTE: The limit as x approaches 2 from the left-hand side would also be 3.
( 4x  5 )  3 .
That is, xlim
 2
2.
lim  ( 2w 2  1)
w3
We need to be able to approach  3 from the left-hand side. Thus, the
2
function y  2w  1 needs to be defined on some open interval of the form
(  3  r ,  3 ) , where r > 0, so that we can approach  3 from the left-hand
2
side. Since the function y  2w  1 is a polynomial, then its domain is the
set of all real numbers. Thus, we can approach  3 from the right-hand side.
2

2
Since as w   3 , 2w  1  19 , then lim  ( 2w  1)  19 .
w3
Answer: 19
Animation of this limit.
NOTE: The limit as w approaches  3 from the right-hand side would also
2
be 19. That is, lim  ( 2w  1)  19 .
w3
3.
lim
t   4
2  5t
t3
We need to be able to approach  4 from the right-hand side. Thus, the
2  5t
y

function
needs to be defined on some open interval of the form
t3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
(  4 ,  4  r ) , where r > 0, so that we can approach  4 from the right-hand
2  5t
side. Since the domain of the function y 
is the set of all real
t3
numbers such that t   3 , then the function is defined on the open interval
(  4 ,  3 ) . Thus, we can approach  4 from the right-hand side. Thus,
lim
t   4
2  5t
22

  22 .
t3
1
Answer:  22
NOTE: The limit as t approaches  4 from the left-hand side would also be
2  5t
lim
  22 .
 22 . That is, t   4
t3
4.
x 2  5 x  24
lim
2
x  3  4 x  7 x  15
We need to be able to approach 3 from the left-hand side. Thus, the function
x 2  5 x  24
y 2
needs to be defined on some open interval of the form
4 x  7 x  15
( 3  r , 3 ) , where r > 0, so that we can approach 3 from the left-hand side.
x 2  5 x  24
Since the domain of the function y 
is the set of all real
4 x 2  7 x  15
5
x


number such that
and x  3 , then the function is defined on the
4
open interval (  1, 3 ) . Thus, we can approach 3 from the left-hand side.
Thus,
( x  3)( x  8)
x 2  5 x  24
x8
11
lim
lim 
lim 

=
=
2

x  3 ( x  3)( 4 x  5)
x  3 4 x  7 x  15
x  3 4x  5
17
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
Answer:
11
17
NOTE: The limit as x approaches 3 from the right-hand side would also be
x 2  5 x  24
11
11

. That is, xlim
.
 3  4 x 2  7 x  15
17
17
5.
lim
x4 
x4
We need to be able to approach 4 from the right-hand side. Thus, the
function y  x  4 needs to be defined on some open interval of the form
( 4 , 4  r ) , where r > 0, so that we can approach 4 from the right-hand side.
Since the domain of the function y  x  4 is the set of real numbers
given by the interval [ 4 ,  ) , then the function is defined on the open interval
( 4 ,  ) . Thus, we can approach 4 from the right-hand side. Thus,
lim
x 4 
x4 
0 0
Answer: 0
6.
lim
x 4 
x4
We need to be able to approach 4 from the left-hand side. Thus, the function
y  x  4 needs to be defined on some open interval of the form
( 4  r , 4 ) , where r > 0, so that we can approach 4 from the left-hand side.
Since the domain of the function y  x  4 is [ 4 ,  ) , then the function is
not defined on any open interval of the form ( 4  r , 4 ) , where r > 0. Thus,
we can NOT approach 4 from the left-hand side. Thus,
lim
x 4 
x  4 = DNE
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
Answer: DNE
Definition A neighborhood of the real number x = a is any open interval
containing the real number a.
Definition A deleted neighborhood of the real number x = a is any open interval
containing the real number a for which the number a is removed (or deleted.)
Theorem Let f be a function defined on a deleted neighborhood of x = a. Then
lim f ( x )  L if and only if lim  f ( x )  L and lim  f ( x )  L .
xa
xa
Example Find xlim
4
xa
x4
This limit is not a one-sided limit. If a side is not specified, then you have to be
able to approach the number from both sides. For the limit to exist, the two onesided limits must exist and be equal. Since lim  x  4  0 by Example 5 above
x4
and x lim
 4
x  4 = DNE by Example 6 above, then lim
x4
x  4 = DNE by the
theorem above.
Answer: DNE
NOTE: The domain of a function plays a role in determining whether we can
approach a number from the right-hand side and/or the left-hand side.
Example Find tlim
5
5t
The domain of the function y 
0  0 and lim 
t 5
5t
is (   , 5 ] . Thus, t lim
 5
5  t = DNE.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
5t =
Answer: DNE
Example Find x lim
4
16  x 2

2
Sign of 16  x :

+

4
The domain of the function y 
Thus,
lim
x4

4
16  x 2 is [  4 , 4 ] .
16  x 2 = DNE. Thus, lim
x4

16  x 2 = DNE
Answer: DNE
16  x 2 = DNE since lim 
x4
NOTE: Also, we have that xlim
4
16  x 2 =
DNE in the example above.
Example Find tlim
7
Sign of
t7
6t  11 :
4
t 7
6t  11

+
+

11

6
The domain of the function y 
Thus, t lim
7
4
4

7
t7
11 

  [7, ) .
is    , 
6t  11
6


t7
lim
6t  11 = DNE. Thus, t  7
4
t 7
6t  11 = DNE
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
Answer: DNE
t7
4
lim
NOTE: Also, we have that
11
6t  11 = DNE since
t
6
lim
 11 
t 

6 


4
t7
6t  11 =
DNE in the example above.
Example Find x lim
 7
Sign of
6
63x 4  9 x 6
63x 4  9 x 6  9 x 4 ( 7  x 2 ) :

+

 7
The domain of the function y 
Thus,
lim 
x  7 
6
6

63 x 4  9 x 6 is 
63x 4  9 x 6 = DNE. Thus, lim
x 7

0
7 ,
6

+

7

7 .
63x 4  9 x 6 = DNE
Answer: DNE
NOTE: Also, we have that
lim 
x   7 
lim
x0
x0
x
7
6
63x 4  9 x 6 = DNE since
63x 4  9 x 6 = DNE in the example above. However,
63x 4  9 x 6  0 since lim 
x0
6
lim 
6
6
lim
6
63x 4  9 x 6  0 and
63x 4  9 x 6  0 .
Another application of One-Sided Limits is to calculate the limit of a piecewise
function at its breakup points.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
Examples Find the following limits, if they exist.
1.
  2x 2 , x  1
f ( x) .
If f ( x )  
, then find xlim
1
5

x
,
x

1

f ( x ) and
We will need to calculate the following one-sided limits xlim
 1
lim f ( x ) . Since the domain of f is the set of all real numbers, we will be
x  1
able to approach 1 from both the right-hand and left-hand sides.
lim f ( x ) = lim  ( 5  x )  5  1  4 . Animation of this right-hand limit.
x 1
x  1
x  1  x  1  f ( x )  5  x
lim  f ( x ) = lim  (  2 x 2 )   2
x 1
x 1
Animation of this left-hand limit.
x  1  x  1  f ( x)   2 x 2
f ( x )  4 and lim  f ( x )   2 . Thus, by the theorem above,
Thus, xlim
1
x 1
lim f ( x ) = DNE.
x 1
Answer: DNE
This graph was created using Maple.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
2.
 x 2  4 x  15 , x   3

g
(
x
)

,  3  x  5 , then find lim g ( x ) .
 2 x  42
If
x3

2
, x5
 3 x  x
We will need to calculate the following one-sided limits
lim
x   3
lim
x   3
g ( x ) and
g ( x ) . Since the domain of g is the set of all real numbers, we will
be able to approach  3 from both the right-hand and left-hand sides.
lim
x   3
g ( x ) = lim 
x3
2 x  42 
x   3  x   3  g ( x ) 
lim
x3

 6  42 
36  6
2x  42
g ( x ) = lim  ( x 2  4 x  15)  9  12  15  6
x3
x   3   x   3  g ( x )  x 2  4 x  15
Thus,
lim
x   3
g ( x )  6 and lim  g ( x )  6 . Thus, by the theorem above,
x3
lim g ( x )  6 .
x3
Answer: 6
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
g ( x ) = DNE since lim  g ( x )   10 and lim  g ( x ) 
NOTE: xlim
5
x5
x5
3.
 3 2
  2 x  12 x  26 , x   2

2x 2  8

h( x )  
, 2x3
x

2


3 14  2 x
, x3


If
, then find
52
lim h( x ) and
x2
lim h( x ) .
x3
We will need to calculate the following one-sided limits
lim
x   2
lim
x   2
h( x ) and
h( x ) . Since the domain of h is the set of all real numbers such that
x  3 , we will be able to approach  2 from both the right-hand and lefthand sides.
lim
2x 2  8
=
x2
lim
h( x ) =
lim
2( x  2)( x  2)
=
x2
x   2
x   2
x   2
lim
lim
x   2
x   2
2( x2  4)
=
x2
2( x  2)   8
2x 2  8
x   2  x   2  h( x ) 
x2
Animation of this right-hand limit.

 3 2

  x  12 x  26    6  24  26   8
x2 
x2
2

3
x   2   x   2  h( x )   x 2  12 x  26
2
Animation of this left-hand limit.
lim

h( x ) =
lim

Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
Thus,
lim
x   2
h( x )   8 and
lim
x   2
h( x )   8 . Thus, by the theorem
h( x )   8 .
above, x lim
2
h( x ) and
We will need to calculate the following one-sided limits xlim
 3
lim h( x ) . Since the domain of h is the set of all real numbers such that
x  3
x  3 , we will be able to approach 3 from both the right-hand and left-hand
sides.
lim h( x ) = lim 
x3
x  3
3
14  2 x 
3
8  2
x  3  x  3  h( x )  3 14  2 x
Animation of this right-hand limit.
2x 2  8
10

 2
lim  h( x ) = lim 
x3
x3
x2
5
2x 2  8
x  3  x  3  h( x ) 
x2
Animation of this left-hand limit.

h( x )  2 and lim  h( x )  2 . Thus, by the theorem above,
Thus, xlim
 3
x3
lim h( x )  2 .
x3
Answer:  8 ; 2
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
4.
lim
x8
x8
x8
Since the absolute value function is a piecewise function, we will need to
x8
y

calculate the one-sided limits. The domain of
is the set of all
x8
real numbers such that x  8 . Thus, we will be able to approach 8 from both
the right-hand and left-hand sides.
lim 
x8
x8
x8
= x lim
8
x8
11
= xlim
8
x8
x  8  x  8  x  8  0  x  8  x  8
lim 
x8
x8
x8
= x lim
8
 ( x  8)
(  1)   1
= x lim
8
x8
x  8   x  8  x  8  0  x  8   ( x  8)
Thus, xlim
8
x8
x8
= DNE
Answer: DNE
COMMENT: In order to find the derivative of the function y  x  8 at x  8 ,
we must calculate the limit above. We will see in a later lesson that the derivative
x8
lim
y

x

8
x

8
of the function
at
is undefined since x  8
= DNE.
x8
5.
25  9t 2
lim
5
3t  5
t
3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
Since the absolute value function is a piecewise function, we will need to
25  9t 2
calculate the one-sided limits. The domain of y  3t  5 is the set of all
5
5
. Thus, we will be able to approach
from
3
3
both the right-hand and left-hand sides.
real numbers such that t 
5
t  
3
lim
 5
t 
 3
lim
 5
t 
 3


 5
t 
 3
lim
 5
t  
 3


 t
5
 3t  5  3t  5  0  3t  5  3t  5
3
25  9t 2
3t  5 =
lim
 5
t 
 3

25  9t 2
3t  5 =
lim
 5
t  
 3

( 5  3t ) ( 5  3t )
=
3t  5
[  ( 3t  5 ) ]   ( 5  5 )   10
5
t  
3
lim


 t
5
 3t  5  3t  5  0  3t  5   ( 3t  5 )  5  3t
3
25  9t 2
3t  5 =
lim
 5
t 
 3

25  9t 2
5  3t =
lim
 5
t  
 3

( 5  3t ) ( 5  3t )
=
5  3t
( 3t  5 )  5  5  10
25  9t 2
Thus, lim5 3t  5 = DNE
t
3
Answer: DNE
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
6.
w 2  11w  28
lim
w7
w7
Since the absolute value function is a piecewise function, we will need to
w 2  11w  28
calculate the one-sided limits. The domain of y 
is the
w7
set of all real numbers such that w   7 . Thus, we will be able to approach
 7 from both the right-hand and left-hand sides.
2
In order to find the absolute value of w  11w  28 , we need to find the
2
sign of w  11w  28 using the material from Lesson 1.
Sign of w2  11w  28  ( w  4 ) (w  7 ) :

+
7


2
Thus, when w   7 , w  11w  28  0 .
2
2
Thus, w  11w  28 =  ( w  11w  28 ) .
Thus,
lim
lim
w  7
w7
w 2  11w  28
w7
 ( w  4 ) (w  7 )
=
w7
=
lim
w7
lim
w7
 ( w 2  11w  28 )
=
w7
[  ( w  4)]  3

2
When w   7 , w  11w  28  0 .
2
2
Thus, w  11w  28 = w  11w  28 .
Thus,
lim
w7
w 2  11w  28
w7
=
lim
w7
w 2  11w  28
=
w7
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
+

4
lim
w7
( w  4 ) (w  7 )
=
w7
Thus, wlim
7
lim
w7
( w  4)   3
w 2  11w  28
w7
= DNE
Answer: DNE
2
COMMENT: In order to find the derivative of the function y  w  11w  28
at w   7 , we must calculate the limit above. We will see in a later lesson that the
2
derivative of the function y  w  11w  28 at w   7 is undefined since
lim
w7
w 2  11w  28
w7
= DNE.
2
What does the graph of y  w  11w  28 look like? You can use the graph of
2
y  w 2  11w  28 to get the graph of y  w  11w  28 . First, graph
y  w 2  11w  28 . Leave the part of the graph that is above the horizontal waxis as it is because this is where the y-coordinates are positive. But since
y  w 2  11w  28 , then this is where w 2  11w  28 is positive and
w 2  11w  28 = w 2  11w  28 when w2  11w  28  y is positive. Leave
the w-intercept(s) as is because this is where the y-coordinate is zero and
w 2  11w  28 = w 2  11w  28 when w2  11w  28  y is zero. Now, for
the part of the graph that is below the horizontal w-axis, the y-coordinates are
2
2
negative. Thus, w  11w  28 =  ( w  11w  28 ) when
w2  11w  28  y is negative. Thus, for the part of the graph that is below the
horizontal w-axis, you reflect it above the horizontal w-axis.
2
In order to graph y  w  11w  28 , we will use the information above that
w 2  11w  28  ( w  4 ) (w  7 ) . Thus, the w-intercepts of the graph are the
points (  7 , 0 ) and (  4 , 0 ) . Using the fact that the vertex of the parabola is on
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
the axis of symmetry and that the axis of symmetry is w  
coordinate of the vertex is w  
11
, then the w2
11
. Since y  ( w  4 ) ( w  7 ) , then the y2
 11   11 8   11 14 
  

 =
coordinate of the parabola is obtained by y      
2  2
2 
 2   2
9
9
 3  3 
 11
,  .
       . Thus, the vertex of the parabola is  
4
4
 2
 2  2 
2
The graph of y  w  11w  28 :
2
The graph of y  w  11w  28 :
These graphs were created using Maple.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
7.
2 x 2  8x
lim
x4
x4
Since the absolute value function is a piecewise function, we will need to
2 x 2  8x
calculate the one-sided limits. The domain of y  x  4
is the set of all
real numbers such that x   4 . Thus, we will be able to approach  4 from
both the right-hand and left-hand sides.
lim
x   4
2 x 2  8x
=
x4
lim
x   4
2 x 2  8x
=
x4
lim
x   4
2x ( x  4 )
=
x4
lim
2x   8
lim
(  2x )  8
x   4
x   4  x   4  x  4  0  x  4  x  4
lim
x   4
2 x 2  8x
=
x4
lim
x   4
2 x 2  8x
=
 ( x  4)
lim
x   4
2x ( x  4 )
=
 ( x  4)
x   4
x   4   x   4  x  4  0  x  4   ( x  4)
2 x 2  8x
Thus, x lim
4
x  4 = DNE
Answer: DNE
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850