l5 infinite limits squeeze theorem
TRANSCRIPT
LIMITSOF
FUNCTIONS
INFINITE LIMITS; VERTICAL AND HORIZONTAL ASYMPTOTES;
SQUEEZE THEOREMOBJECTIVES:•define infinite limits;•illustrate the infinite limits ; and•use the theorems to evaluate the limits of functions.•determine vertical and horizontal asymptotes•define squeeze theorem
DEFINITION: INFINITE LIMITS
Sometimes one-sided or two-sided limits fail to exist because the value of the function increase or decrease without bound. For example, consider the behavior of forvalues of x near 0. It is evident from the table and graph in Fig 1.1.15 that as x values are taken closer and closer to 0 from the right, the values ofare positive and increase without bound; and asx-values are taken closer and closer to 0 from the left, the values of are negative and decrease without bound.
x
1)x(f =
x
1)x(f =
x
1)x(f =
In symbols, we write
−∞=+∞=−→→ + x
1lim and
x
1lim
0x0x
Note:The symbols here are not real numbers; they simply describe particular ways in which the limits fail to exist. Thus it is incorrect to write .
∞−∞+ and
( ) ( ) 0=∞+−∞+
Figure 1.1.15 (p. 74)
1.1.4 (p. 75) Infinite Limits (An Informal View)
Figure 1.2.2 (p. 84)
Figure 1.1.2 illustrate graphically the limits for rational functions of the form .
( ) ( ) ( ) 22 ax
1 ,
ax
1 ,
ax
1
−−
−−
EXAMPLE: Evaluate the following limits:
40x x
1 lim .1
+→
40x x
1 lim .2
−→
50x x
1 lim .4
+→
−∞==−→ − 0
1
x
1 lim
50x
+∞==+→ + 0
1
x
1 lim
40x
+∞==+→ − 0
1
x
1 lim
40x
+∞==+→ + 0
1
x
1 lim
50x
50x x
1 lim .5
−→
+∞=→ 40x x
1 lim .3 ∞=
→ 50x x
1 lim .6
2x
x3 lim a. .7
2x −−→
2x
x3 lim.b
2x −+→
( )( ) ∞−−+=+==
→ 0
6
0
23
2-x
3x lim
-2x
2.028.12x
8.1 say ,left from2 to close x of value
take we means2x
−=−=−
→ −
( )( ) +∞=++=+==
+→ 0
6
0
23
2-x
3x lim
2x
1.021.22x
1.2 say ,right from2 to close x of value
take we means2x
+=−=−
→ +
∞=−→ 2x
x3 lim .c
2x
)x(flimax +→
)x(flimax −→
)x(flimax→
∞+ ∞+ ∞+
∞+
∞+
∞− ∞− ∞−∞−
∞−
∞
∞
SUMMARY:
)x(Q
)x(R)x(f If =
EXAMPLE
( ) +∞=
+∞+=
++
−
=+
+∞=−
+
++
→
→→
3
1
3x
2
3x
2 lim ,then
3
1
3x
2 lim and
3x
2 lim .1
3x
3x3x
( )( ) −∞=+∞−
+∞=+∞+c
c
nSubtractio/Addition:Note
( ) ( ) ( ) ( )( ) ( ) ( ) ( ) +∞=−∞−−∞=+∞−
−∞=−∞++∞=+∞+c c
c c
:Note
( ) ( )( ) −∞=−∞+=
+−⋅
−
−=+−+∞=
−
+
++
→
→→
11x
3x
1x
x2 lim ,then
11x
3x lim and
1x
x2 lim .2
1x
1x1x
( ) +∞=
−∞−=
+−⋅
−
−=+−−∞=
−
−
−−
→
→→
3
1
4x
6x2
2x
x3 lim ,then
3
1
4x
6x2 lim and
2x
x3 lim .3
2x
2x2x
VERTICAL AND HORIZONTAL ASYMPTOTES
DEFINITION:
−∞=
+∞=
−∞=
+∞=
−
−
+
+
→
→
→
→
)x(flim.d
)x(flim.c
)x(flim.b
)x(flim.a
ax
ax
ax
ax
The line is a vertical asymptote of the graph of the function if at least one of the following statement is true:
x a=( )y f x=
x=a
0
+∞=+→
)x(flimax
+∞=−→
)x(flimax
The following figures illustrate the vertical asymptote . x a=
x=a
0
x=a
0
x=a
−∞=−→
)x(flimax
−∞=+→
)x(flimax
The following figures illustrate the vertical asymptote . x a=
0
DEFINITION:
b)x(flim or b)x(flimxx
==−∞→+∞→
The line is a horizontal asymptote of the graph of the function if either
by =( )y f x=
y=b
0
y=b
b)x(flimx
=+∞→
The following figures illustrate the horizontal asymptote
by =
0
b)x(flimx
=+∞→
y=b
0
y=b
b)x(flim x
=∞−→
The following figures illustrate the horizontal asymptote by =
0
b)x(flimx
=−∞→
Determine the horizontal and vertical asymptote of the function and sketch the graph.( ) 3
2f x
x=
−
a. Vertical Asymptote: Equate the denominator to zero to solve for the vertical asymptote.
2x02x =⇒=−Evaluate the limit as x approaches 2
2
3 3 3lim
2 2 2 0x x→= = = ∞
− −
b. Horizontal Asymptote:Divide both the numerator and the denominator by the highest power of x to solve for the horizontal asymptote.
3 30
lim 02 2 1 01
x
xxx x
→+∞
+∞= = =−− −
+∞
3 30
lim 02 2 1 01
x
xxx x
→−∞
−∞= = =−− −
−∞
∴
( )
erceptintx no is there therefore
30 ;2x
30 ,0)xf( If
2
3
20
3xf ,0x If
:Intercepts
−
≠−
==
−=−
==
.asymptote horizontal a is 0 ,Thus
•
−
2
3,0
VA: x=2
HA:y=00
( ) 3
2f x
x=
−
Determine the horizontal and vertical asymptote of the function and sketch the graph.( )
3x
1x2xf
−+=
a. Vertical Asymptote: b. Horizontal Asymptote:
3x03x =⇒=−
∞==−+
→ 0
7
3x
1x2lim
3x
21
2
x3
xx
x1
xx2
limx
==−
+
∞→
asymptote horizontal a is 2y =∴asymptote vertical a is 3x =∴
( )
2
1x ;
3x
1x20 ,0)xf( If
3
1
30
10xf ,0x If
:Intercepts
−=−+==
−=−+==
HA:y=2
VA:x=3
o
( )3x
1x2xf
−+=
SQUEEZE THEOREM
LIMITS OF FUNCTIONS USING THE SQUEEZE PRINCIPLE
The Squeeze Principle is used on limit problems where the usual algebraic methods (factoring, conjugation, algebraic manipulation, etc.) are not effective. However, it requires that you be able to ``squeeze'' your problem in between two other ``simpler'' functions whose limits are easily computable and equal. The use of the Squeeze Principle requires accurate analysis, algebra skills, and careful use of inequalities. The method of squeezing is used to prove that f(x)→L as x→c by “trapping or squeezing” f between two functions, g and h, whose limits as x→c are known with certainty to be L.
SQUEEZE PRINCIPLE :
Lf(x)lim then
h(x)lim Lg(x)lim and
h(x)f(x)g(x) satisfy h and ,g ,f functions that Assume
ax
axax
=
==
≤≤
→
→→
Theorem 1.6.5 (p. 123)
Figure 1.6.3 (p. 123)
EXAMPLE:
x
cos4x-cos3x-2lim 4.
x5sin
x3sinlim .3
x
x2sinlim .2
x
xtanlim .1
limits. following the Evaluate
0x0x
0x0x
→→
→→
( ) ( ) 111
xcos
1lim
x
xsinlim
xcos
1
x
xsin lim
x
xtanlim .1
0x0x
0x0x
==
=
•=
→→
→→
( ) ( ) 212 x2
x2sinlim2
2
2
x
x2sin lim
x
x2sinlim .2
0x
0x0x
==
=
•=
→
→→
SOLUTION:
5
3
15
13
x5x5sin
5
x3x3sin
3lim
xx5sin
xx3sin
limx5sin
x3sinlim .3
0x
0x0x
=••=
•
•=
=
→
→→
( )
( ) ( ) 00403
4x
cos4x-1lim4
3x
cos3x-1lim3
x
cos4x-1lim
x
cos3x-1lim
x
x4cosx3cos11 lim
x
cos4x-cos3x-2lim 4.
0x0x
0x0x
0x
0x
=•+•=
+
=
+
=
−−+=
→→
→→
→
→
3x2x
2xx lim .4
4x
x16 lim .3
4t
2t lim .2
x9
x4 lim .1
2
2
3x
2
4x
22t
2
2
3x
−−++
−−
−+
−
→
→
→
→
−
+
EXERCISES: Evaluate the following limits: