LIMITS OF FUNCTIONS
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LIMITSOF
FUNCTIONS
LIMITS OF FUNCTIONS
OBJECTIVES:•define limits;•illustrate limits and its theorems; and•evaluate limits applying the given theorems.• define one-sided limits• illustrate one-sided limits• investigate the limit if it exist or not using
the concept of one-sided limits.•define limits at infinity;•illustrate the limits at infinity; and•determine the horizontal asymptote.
DEFINITION: LIMITS The most basic use of limits is to describe how a function behaves as the independent variable approaches a given value. For example let us examine the behavior of the function for x-values closer and closer to 2. It is evident from the graph and the table in the next slide that the values of f(x) get closer and closer to 3 as the values of x are selected closer and closer to 2 on either the left or right side of 2. We describe this by saying that the “limit of is 3 as x approaches 2 from either side,” we write
1xx)x(f 2 +−=
1xx)x(f 2 +−=
( ) 31xxlim 2
2x=+−
→
2
3
f(x)
f(x)
x
y
1xxy 2 +−=
x 1.9 1.95 1.99 1.995 1.999 2 2.001 2.005 2.01 2.05 2.1
F(x) 2.71 2.852 2.97 2.985 2.997 3.003 3.015 3.031 3.152 3.31
left side right side
O
1.1.1 (p. 70) Limits (An Informal View)
This leads us to the following general idea.
EXAMPLE
Use numerical evidence to make a conjecture about the value of .
1x
1xlim
1x −−
→
Although the function is undefined at x=1, this has no bearing on the limit. The table shows sample x-values approaching 1 from the left side and from the right side. In both cases the corresponding values of f(x) appear to get closer and closer to 2, and hence we conjecture that and is consistent with the graph of f.
1x
1x)x(f
−−=
21x
1xlim
1x=
−−
→
Figure 1.1.9 (p. 71)
x .99 .999 .9999 .99999 1 1.00001 1.0001 1.001 1.01
F(x) 1.9949 1.9995 1.99995 1.999995 2.000005 2.00005 2.0005 2.004915
THEOREMS ON LIMITS
Our strategy for finding limits algebraically has two parts:•First we will obtain the limits of some simpler function•Then we will develop a list of theorems that will enable us to use the limits of simple functions as building blocks for finding limits of more complicated functions.
We start with the following basic theorems, which are illustrated in Fig 1.2.1
Theorem 1.2.1 (p. 80)
( ) ( ) axlim b kklim a
numbers. real be k and a Let Theorem 1.2.1
axax==
→→
Figure 1.2.1 (p. 80)
( )
33lim 33lim 33lim
example, For
a. of values all for ax as kf(x)
why explains whichvaries, x as k at fixed remain
f(x) of values the then function, constant a is k xf If
x0x-25x===
→→
=
→→→ π
Example 1.
( ) ( )
ππ
=−==
→→=
→→→xlim 2xlim 0xlim
example, For
. axf that true be also must it ax then x, xf If
x-2x0x
Example 2.
Theorem 1.2.2 (p. 81)
The following theorem will be our basic tool for finding limits algebraically.
This theorem can be stated informally as follows:
a) The limit of a sum is the sum of the limits.b) The limit of a difference is the difference of the limits.c) The limits of a product is the product of the limits.d)The limits of a quotient is the quotient of the limits, provided the limit of the denominator is not zero.e) The limit of the nth root is the nth root of the limit.
•A constant factor can be moved through a limit symbol.
( )5x2lim .14x
+→
( )12x6lim .23x
−→
( ) )2x5(x4lim .33x
−−→
EXAMPLE : Evaluate the following limits.
31
58
5)4(2
5limxlim2
5limx2lim
4x4x
4x4x
=+=
+=
+=
+=
→→
→→( )
6
12-18
12)3(6
12limx6lim3x3x
==
−=
−=→→
( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )13
131
2)3(534
2limxlim5xlim4lim
2limx5limxlim4lim
2x5limx4lim
3x3x3x3x
3x3x3x3x
3x3x
==
−−=
−⋅−=
−⋅−=
−•−=
→→→→
→→→→
→→
4x5
x2lim .4
5x −→
( ) 3
3x6x3lim .5 +
→
3x
1x8lim .6
1x ++
→
( )21
10
425
52 =−
=
( ) ( ) ( ) ( )4limxlim5
x lim2
4limx5lim
x2 lim
5x5x
5x
5x5x
5x
→→
→
→→
→
−=
−=
( )( )( )
( )( )( ) ( )
3375
15633
6limxlim3
6limx3lim
6x3lim
33
3
3x3x
3
3x3x
3
3x
==+⋅=
+=
+=
+=
→→
→→
→
2
3
4
9
3x
1x8lim
1x
==
++=
→
OR
When evaluating the limit of a function at a given value, simply replace the variable by the indicated limit then solve for the value of the function:
( ) ( ) ( )
22
3lim 3 4 1 3 3 4 3 1
27 12 1
38
xx x
→+ − = + −
= + −=
EXAMPLE: Evaluate the following limits.
2x
8xlim .1
3
2x ++
−→
Solution:( )
0
0
0
88
22
82
2x
8xlim
33
2x=+−=
+−+−=
++
−→
Equivalent function:
(indeterminate)
( ) ( )2x
4x2x2xlim
2
2x ++−+=
−→
( )( ) ( )
12444
4222
4x2xlim
2
2
2x
=++=+−−−=
+−=−→
122x
8xlim
3
2x=
++∴
−→
Note: In evaluating a limit of a quotient which reduces to , simplify the fraction. Just remove
the common factor in the numerator and denominator which makes the quotient .
To do this use factoring or rationalizing the numerator or denominator, wherever the radical is.
0
0
0
0
x
22xlim .2
0x
−+→
Solution:
Rationalizing the numerator:
(indeterminate)0
0
0
220
x
22xlim
0x=−+=−+
→
( )22xx
22xlim
22x
22x
x
22xlim
0x0x ++−+=
++++•−+=
→→
( ) 4
2
22
1
22
1
22x
1lim
22xx
xlim
0x0x==
+=
++=
++=
→→
4
2
x
22xlim
0x=−+∴
→
9x4
27x8lim .3
2
3
2
3x −
−→
Solution:
By Factoring:
(indeterminate)32
3
3
22
38 27
8 27 27 27 02lim
4 9 9 9 034 9
2
x
x
x→
− ÷− − = = =− − − ÷
( ) ( )( ) ( ) ( )
+
+
+
=+
++=−+
++−=→→ 3
23
2
923
623
4
3x2
9x6x4lim
3x23x2
9x6x43x2lim
2
2
2
3x
2
2
3x
2
23
2
3
2
9
6
27
33
999 ====+
++=
2
23
9x4
27x8lim
2
3
2
3x
=−
−∴→
5x
3x2xlim .4
2
3
2x +++
→
Solution:
( ) ( )( )
33
222
2 2 2 32 3lim
5 2 5
8 4 3
4 5
15
9
15
3
x
x x
x→
+ ++ + =+ +
+ +=+
=
=
3
15
5x
3x2xlim
2
3
2x=
+++∴
→
DEFINITION: One-Sided Limits
The limit of a function is called two-sided limit if it requires the values of f(x) to get closer and closer to a number as the values of x are taken from either side of x=a. However some functions exhibit different behaviors on the two sides of an x-value a in which case it is necessary to distinguish whether the values of x near a are on the left side or on the right side of a for purposes of investigating limiting behavior.
Consider the function
<−>
==0x ,1
0x ,1
x
x)x(f
1
-1
As x approaches 0 from the right, the values of f(x) approach a limit of 1, and similarly , as x approaches 0 from the left, the values of f(x) approach a limit of -1.
1x
xlim and 1
x
xlim
,symbols In
oxox−==
−+ →→
1.1.2 (p. 72) One-Sided Limits (An Informal View)
This leads to the general idea of a one-sided limit
1.1.3 (p. 73) The Relationship Between One-Sided and Two-Sided Limits
EXAMPLE:
x
x)x(f =1. Find if the two sided limits exist given
1
-1
exist. not does x
xlim or
exist not does itlim sided two the thenx
xlim
x
xlim the cesin
1x
xlim and 1
x
xlim
ox
oxox
oxox
→
→→
→→
−+
−+
≠
−==
SOLUTION
EXAMPLE:2. For the functions in Fig 1.1.13, find the one-sided limit and the two-sided limits at x=a if they exists.
The functions in all three figures have the same one-sided limits as , since the functions are Identical, except at x=a.
ax →
1)x(flim and 3)x(flim
are itslim These
axax==
−+ →→
In all three cases the two-sided limit does not exist as because the one sided limits are not equal. ax →
SOLUTION
Figure 1.1.13 (p. 73)
3. Find if the two-sided limit exists and sketch the graph of
2
6+x if x < -2( ) =
x if x -2g x
≥
( )
4
26
x6lim)x(glim.a2x2x
=−=
+=−− −→−→
( )4
2-
xlim)x(glim.b
2
2
2x2x
==
=++ −→−→
4)x(glim or
4 to equal is and exist itlim sided two the then
)x(glim)x(glim the cesin
2x
2x2x
=
=
−→
−→−→ −+
SOLUTION
EXAMPLE:
x-2-6 4
y
4
4. Find if the two-sided limit exists and sketch the graph of and sketch the graph.
2
2
3 + x if x < -2
( ) = 0 if x = -2
11 - x if x > -2
f x
SOLUTION
( )( )
7
23
x3lim)x(flim.a
2
2
2x2x
=−+=
+=−− −→−→
( )( )
7
2-11
x11lim)x(flim.b
2
2
2x2x
=−=
−=++ −→−→
7)x(flim or
7 to equal is and exist itlim sided two the then
)x(flim)x(flimthe cesin
2x
2x2x
=
=
−→
−→−→ −+
EXAMPLE:
graph. the sketch and
,exist f(x) lim if eminerdet ,4x23)x(f If .52x→
−+=
( )3
4223
4x23lim )x(flim .a2x2x
=−+=
−+=−− →→
( )3
4223
4x23lim )x(flim .b2x2x
=−+=
−+=++ →→
3)x(flim or
3 to equal is and exist itlimsided two the then
)x(flim)x(flimthe cesin
2x
2x2x
=
=
→
→→ −+
SOLUTION
EXAMPLE:
f(x)
x
(2,3)
2
DEFINITION: LIMITS AT INFINITY
The behavior of a function as x increases or decreases without bound is sometimes called the end behavior of the function.
)x(f
If the values of the variable x increase without bound, then we write , and if the values of x decrease without bound, then we write .
+∞→x−∞→x
For example ,
0x
1lim and 0
x
1lim
xx==
+∞→−∞→
x
x
0x
1limx
=−∞→
0x
1limx
=+∞→
1.3.1 (p. 89) Limits at Infinity (An Informal View)
In general, we will use the following notation.
Figure 1.3.2 (p. 89)
Fig.1.3.2 illustrates the end behavior of the function f when L)x(flim or L)x(flim
xx==
−∞→+∞→
Figure 1.3.4 (p. 90)
EXAMPLEFig.1.3.2 illustrates the graph of . As suggested by this graph,
x
x
11y
+=
ex
11lim
and ex
11lim
x
x
x
x
=
+
=
+
−∞→
+∞→
EXAMPLE ( Examples 7-11 from pages 92-95)
6x32x
lim .4
x311x2x5
lim .3
5x2xx4
lim .2
8x65x3
lim .1
2
x
23
x
3
2
x
x
−+−
+−−−
−+
+∞→
+∞→
−∞→
+∞→( )
( )336
x
36
x
xx5xlim .6
x5xlim .5
−+
−+
+∞→
+∞→
EXERCISES:
( )
( ) ( )
( )
5w4w
7w7wlim 10.
2x
8xlim .5
19x9x2lim 9. 4y
y8y4lim .4
1y2y
3y2y1ylim 8.
1x
4x3xlim .3
1x
3x2x3x2lim 7.
4x3x
1x2lim .2
1x9
1x3lim 6. 2x5x4lim .1
2
2
1w
3
2x
2
134
5x
3
13
2y
2
2
1y3
2
1x
2
23
1x21x
2
3
1x
2
3x
−−++
−−
+−
++
+−−+−
+++
−−−+
+−+
−−+−
−→→
−
→→
→−→
→−→
→→
A. Evaluate the following limits.
EXERCISES:
B. Sketch the graph of the following functions and the indicated limit if it exists. find
.
)x(glim.c g(x) lim.b g(x)lim.a
1x if 2x-7
1x if 2
1x if 3x2
)x(g .2
)x(flim.c f(x) lim.b f(x)lim.a
4- x if 4x
-4x if x4)x(f.1
1x1x1x
4x4x4x
→→→
−→−→−→
−+
−+
>=<+
=
≤+>−
=
.
)x(flim.c f(x) lim.b f(x)lim.a
1x2)x(g .5
)x(flim.c f(x) lim.b f(x)lim.a
x4)x(g .4
)x(flim.c f(x) lim.b f(x)lim.a
0x if 3
0x if x)x(f.3
2
1x
2
1x
2
1x
4x4x4x
0x0x0x
→→→
→→→
→→→
−+
−+
−+
−=
−=
=≠
=
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