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MTH 104. Calculus and Analytical Geometry. Lecture # 11. L’HÔPITAL’S RULE: INDETERMINATE FORMS. INDETERMINATE FORMS OF TYPE. Limit of the form. in which. and. is called an indeterminate form of type. L’Hopital Rule for form 0/0. Suppose that. and. - PowerPoint PPT Presentation
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Calculus and Analytical Geometry
Lecture # 11
MTH 104
L’HÔPITAL’S RULE: INDETERMINATE FORMS
INDETERMINATE FORMS OF TYPE
0 0
( )lim
( )x a
f x
g x
( ) 0f x x a
0 0
Limit of the form
in which and
is called an indeterminate form of type
.
( ) 0 as g x
L’Hopital Rule for form 0/0
f g
x a x a
lim ( ) 0 lim ( ) 0x a x a
f x and g x
Suppose that and are differentiable functions on an open
, except possible at , and that
interval containing
lim ( ) / ( )x a
f x g x
If exists, or if this limit is or , then
( )lim
( )x a
f x
g x
( )lim
( )x a
f x
g x
Moreover this statement is also true in the case of limits as
, , ,x a x a x ,x or as
EXAMPLE Find the limit 2
2
4lim
2x
x
x
Using L’HÔPITAL’S rule, and check the result by factoring.
solution2 2
2
4 2 4lim
2 2 2x
x
x
0
0
form
Using L’HÔPITAL’S rule
2
2
4lim
2x
dx
dxdx
dx
2
2
4lim
2x
x
x
2
2lim
1x
x
4
By computation2
2
4lim
2x
x
x
2
( 2)( 2)lim
( 2)x
x x
x
4 2
lim 2x
x
ExampleIn each part confirm that the limit is an indeterminate form of type 0/0 and evaluate it using L’HOPITAL’s rule
0
sin 2limx
x
x / 2
1 sinlim
cosx
x
x
(a)
(b)
30
1lim
x
x
e
x
2
0
tanlimx
x
x
(c)
(d)
20
1 coslimx
x
x
4 / 3lim
sin(1/ )x
x
x
(e)
(f)
Solution
(a)0
sin 2limx
x
x
sin 0
0%
Applying L’HÔPITAL’S rule
0
sin 2limx
x
x
0
(sin 2 )lim
( )x
dx
dxdx
dx
0
2cos 2lim 2cos(0)
1x
x
2
(b)
/ 2
1 sinlim
cosx
x
x
1 sin 2cos 2
1 1 000
Applying L’HÔPITAL’S rule
/ 2
1 sinlim
cosx
x
x
/ 2
1 sinlim
cosx
dx
dxd
xdx
/ 2
coslim
sinx
x
x
cos 2sin 2
00
1
(c)30
1lim
x
x
e
x
0 1
0
e 1 1
0
0
0
Applying L’HÔPITAL’S rule
30
1lim
x
x
e
x
20lim
3
x
x
e
x
0
0
e 1
0
20
tanlimx
x
x
tan 0
0 0
0
(d)
Applying L’HÔPITAL’S rule
20
tanlimx
x
x
20
tanlimx
dx
dxdx
dx
2
0
seclim
2x
x
x
INDETERMINATE FORMS OF TYPE
( ) / ( )f x g x
The Limit of a ratio, in which the numerator has limit and the denominator has the limit is called an indetreminate form of type L’Hopital Rule for form Suppose f and g are differentiable functions on an open interval conatining x=a, except possibly at, x=a and that
lim ( )x a
f x
and lim ( )x a
g x
lim ( ) ( )x a
f x g x
If exists, or if this limit is or , then
( ) ( )lim lim
( ) ( )x a x a
f x f x
g x g x
Moreover this statement is also true in the case of limits as
, , ,x a x a x ,x or as
Example
In each part confirm that the limit is indeterminate form of type
and apply L’HÔPITAL’S
limxx
x
e 0
lnlim
cscx
x
x(a) (b)
solution
limxx
x
e e
(a)
Applying L’HÔPITAL’S rule
limxx
x
e
( )lim
( )x x
dx
dxde
dx
1lim
xx e 1 0
0
lnlim
cscx
x
x
ln 0
csc(0)
(b)
Applying L’HÔPITAL’S rule
0
lnlim
cscx
x
x
0
(ln )lim
(csc )x
dx
dxd
xdx
0
1lim
csc cotx
x
x x
1 x csc x cot xAny additional application of L’HÔPITAL’S rule will yield powers of
in the numerator and expressions involving and
in the denominator.
Rewriting last expression
0
sinlim tanx
xx
x
0
sinlimx
x
x
0lim tanx
x
(1)(0) 0
Thus, 0
lnlim 0
cscx
x
x
INDETERMINATE FORMS OF TYPE0.
The limit of an expression that has one of the forms
( )( ), ( ). ( ), ( ) , ( ) ( ), ( ) ( )
( )g xf x
f x g x f x f x g x f x g xg x
( )f x ( )g xis called and indeterminate form if the limits andindividually exert conflicting influences on the limit of the entire expression.
For example
0lim ln 0.x
x x
On the other hand
2lim 1x
x x
( )
Indeterminate form
Not an indeterminate form
0.
0 / 0
Indeterminate form of type
can sometimes be evaluated by rewriting the product as a ratio, and then applying L’HÔPITAL’S rule for indeterminate form of type or .
Example Evaluate
0lim lnx
x x
(a) (b)/ 4
lim (1 tan )sec2x
x x
0lim lnx
x x
0.( )
Solution
(a)
0lim lnx
x x
0
lnlim
1x
x
x
Rewriting form
Applying L’HÔPITAL’S rule
0lim lnx
x x
0
lnlim
1x
x
x
0
(ln )lim
1( )x
dx
dxd
xdx
0 2
1lim
1x
x
x
2
0limx
x
x
0lim ( )x
x
0
/ 4lim (1 tan )sec2x
x x
(b)
0
Rewriting as
4
lim 1 tan sec2x
x x
4
1 tanlim
1 sec2x
x
x
4
1 tanlim
cos 2x
x
x
0
0
Applying L’HÔPITAL’S rule
4
lim 1 tan sec2x
x x
4
1 tanlim
cos 2x
x
x
2
4
seclim
2sin 2x
x
x
2(sec )42
2sin4
2
12
4
1 tanlim
cos2x
dx
dxd
xdx
Indeterminate forms of type
A limit problem that leads to one of the expressions
,
,
is called an indeterminate form type
.
The limit problems that lead to one of the expressions
,
,
are not indeterminate, since two terms work together.
Example Evaluate
0
1 1lim
sinx x x
Solution
0
1 1lim
sinx x x
1 10 sin 0
0
1 1lim
sinx x x
0
sin
sinlimx
x x
x x
sin 0 00sin 0
Rewriting
00
Applying L’HÔPITAL’S rule
0
1 1
sinlimx x x
0
sin
sinlimx
x x
x x
0
sin
sinlimx
dx x
dxdx x
dx
0
cos 1
sin coslimx
x
x x x
0
0
Again Applying L’HÔPITAL’S rule
0
cos 1lim
sin cosx
dx
dxd
x x xdx
0
1 1
sinlimx x x
0
sinlim
cos sin cosx
x
x x x x
sin 0
cos0 0sin 0 cos0
0
01 0 1
INDETERMINATE FORM OF TYPE 0 00 , ,1
( )lim ( ) xf x g
can give rise to indeterminate forms of the types and 0 00 , 1
1
0lim (1 ) x
xx
It is indeterminate because the expressions and gives --Two conflicting influences. Such inderminate form can be evaluated by first introducing a dependent variable
( )( )g xy f x
1 and respectively.
( )1n 1n ( )
g xy f x
( ) 1n ( )g x f x
The limit of lny will be an indeterminate form of type 0
Limits of the form
For example
(1 form)
1 x1x
Example1
0lim (1 ) xx
x e
Solution Let 1(1 ) xy x
1
ln ln(1 ) xy x 1
ln ln(1 )y xx
0 0
ln(1 )lim ln limx x
xy
x
0( form)0ln(1 0)
0
Applying L’HÔPITAL’S rule
0 0
1limln lim 1
1x xy
x
0 0
11lim ln lim1x x
xy
ln 1 as 0y x
ln 1 as 0ye e x as 0.y e x
1
0Thus, lim 1 .x
xx e