Transcript Example: Find

羅必達法則
(L’Hospital’s Rule)
1.不定式 (Indeterminate Forms)
2.羅必達定理(L’Hopital’s Rule)
3. 例題
page 659-663
1
Indeterminate Forms
0
0


1. The Indeterminate Forms of Type
2. The Indeterminate Forms of Type
3. The Indeterminate Forms 0   and 
4. The Indeterminate Forms 00 , 1 and  0
EX: lim
x 0
2
1  cos x
x
,
,
lim
2
x
x  3 x x  e
lim
x 0 
( x  1)cos x , lim
x 
ln x
sin(3x)
, lim
x x 0
x
The Indeterminate Forms
of Type 0/0
x2  1
Take lim
for example
x 1 x  1
2
x
1  0 &
When x  1  lim
x1
limx  1  0
x1
x2  1 0

lim
 x1 x  1 0
Divide both numerator and denominator by x-1
x2  1
2
lim
x 1
x 1
3
x 1
lim
1
x 1 x  1
The Indeterminate Forms
of Type 0/0
x2  1
 lim
lim
x 1
x 1
x 1
x2  1
x 1 
x 1
x 1
x2  1
lim
x 1 2
x 1

x 1 1
lim
x 1 x  1
x2  1
2
2
lim
(
x
 1)|
2 x x 1 2
x 1
x

1
x 1
x 0



2
lim
x

1

x 1
( x  1) x 0
1 x 1 1
x 1
lim
x 1 x  1
4
The Indeterminate Forms
of Type 0/0
Replace x 2  1 by f ( x)  f (u), f (u)  0
Replace x  1 by g ( x)  g (u), g (u)  0
Replace x −1 by x  u
if
f  x   f (u )
g  x   g (u )
lim f  x  lim g  x  lim
,lim
x
, x
, xu x  u xu x  u
exist and g (u)  0 , then the weak form of
L’Hopital’s Rule
lim
x u
5
f ( x)
f ' ( x)

g ( x) g ' ( x)
L’Hospital’s Rule
Let f and g be functions and let a be a real
number such that
lim f  x   0,lim g  x   0
xa
xa
Let f and g have derivative that exist at each
point fin' xsome
open interval containing a

f  x
lim
L
lim
L
g ' x 
If
, then
g  x
f ' x 
f ' x 
lim
If
does exist because g ' x 
g ' x 
becomes large without bound for values of x
near a, then lim f  x  also does not exist
xa
x a
x a
x u
6
g (u )
EX1 L’Hospital’s Rule
x2  1
lim
x 1
x 1
Find
Check the conditions of L’Hospital’s Rule
2
x
lim  1  0
x1
limx  1  0
x1
x2  1 0

lim
x

1
0
x 1
If f  x   x 2  1
then f’(x)=2x
If f(x)=x-1
then f’(x)=1
By L’Hospital’s Rule, this result is the desired limit:
x 2  1 '

x2  1
2x
 lim
 lim
2
lim
x 1
1
x 1
x 1  x  1 '
x 1
7
EX2 L’Hospital’s Rule
ln x
2
Find lim
x 1  x  1
Check the conditions of L’Hospital’s Rule
lim ln x  0
x1
If
If
lim x  1
x1
2
0
ln x
lim  x  1
x1
2

0
0
then f’(x)= x 1
f(x)=  x  12 then f’(x)=2(x-1)
f  x   ln x
x 1
Because lim
does not exist
x1 2  x  1
ln x
Then
does not exist
2
lim
x 1  x  1
8
Using L’Hospital’s Rule
f  x  g  x  leads to the
1. Be sure that lim
xa
indeterminate form 0/0.
2. Take the derivates of f and g seperately.
f '  x  g '  x  ; this limit, if it
3. Find the limit of lim
xa
exists, equals the limit of f(x)/g(x).
4. If necessary, apply L’Hospital’s rule more
than once.
9
EX3 L’Hospital’s Rule
x3
lim
x
x 0 e  1
Find
Check the conditions of L’Hospital’s Rule
limx
3
lim  e
0
x 1
x 1
If
If
f  x   x3
f(x)= e x  1
2
lim3x 2
x
 1  0
then f’(x)= 3x 2
then f’(x)= e x
3x
0
x0


0
lim
x
x
lim e
1
x 0 e
x0
10
x3
0

lim
x
0
x 1 e  1
x3
0
lim
x
x 1 e  1
EX4-1 L’Hospital’s Rule
ex  x  x
lim
x2
x 0
Find
lim  e
x
x 0
 x  1  0 lim
x0
x
e
x  0 lim  x2  1  0
x
0
x 0
2
If f  x   e x  x  1 then f’(x)= e x  1
If f(x)= x 2
then f’(x)=2x

ex 1 0

lim
2x
0
x 0
11
EX4-2 L’Hospital’s Rule
ex  x 1
ex 1 0
 lim

lim
2
x 0
x
2
x
0
x 0
x
f
x

e
1


If
If f(x)= 2x
then f’(x)= e x
then f’(x)=2
ex 1

lim
2
x 0 2
ex  x 1
ex 1
ex 1
 lim
 lim 
lim
2
x

0
x 0 2
x
2x
2
x 0
12
EX5 L’Hospital’s Rule
x2  1
lim
x
x 1
Find
lim  x
x 1

2
 1  0
ex  x 1 0 0
 
lim
1 0
x
x 1
x 1
lim
x 1
x2  1
2x

1
lim
lim
 12
1
x
x1
x1 2 x
x 1

lim
x
x 1
2
lim  x
x 1
lim
x 1
13
2
 1
x

0
0
1
(by substitution)
Proof of L’Hospital’s Rule-1
We can prove the theorem for special case f, g,
f’, g’ are continuous on some open interval
containing a, and g’(a)=0. With these
assumptions the fact that
lim f  x   0 and lim g  x   0
xa
xa
means that both
f(a)=0 and g(a)=0
14
Proof of L’Hospital’s Rule-2
Thus,
f  x
f  x   f (a )
lim
 lim
x a g ( x )
x a g ( x )  g ( a )
Multiplying the numerator and denominator
by 1/(x-a) gives
f  x   f (a )
f  x
xa
lim
 lim
x a g ( x )
x a g ( x )  g ( a )
xa
15
Proof of L’Hospital’s Rule-3
By the property of limits, this becomes,
f  x   f (a )
f  x  lim
x a
xa
lim

x a g ( x )
g ( x)  g (a)
lim
x a
xa
the limit of numerator is f’(a)
the limit of denominator is g’(a) and g (a)  0
f ' x 
f '  a  lim
f ' x 
x a

 lim
g '(a) lim g '( x) xa g '( x)
x a
16
Proof of L’Hospital’s Rule-4
Thus,
f  x
f ' x 
f ''  x 
lim
 lim
 lim

x a g ( x )
x a g '( x )
x a g ''( x )
17
sin x  x
.(0/0)
Example: Find lim
3
x 0
x
lim(sin x  1)  0, lim x  0
3
x 0
x 0
sin x  x
cos x  1
 sin x
lim
 lim
 lim
3
2
x 0
x 0
x 0
x
3x
6x
 cos x  1
 lim

x 0
6
6
18
1  cos x
Example: Find lim
(0/0)
x 0 x 2  3 x
lim(1  cos x)  0, lim  x 2  3x   0
x 0
x 0
1  cos x
sin x
lim 2
 lim
0
x 0 x  3 x
x 0 2 x  3
19
Example: Find
e x  ln(1  x)  1
lim
x 0
x2
(0/0)
lim(e x  ln(1  x)  1)  0, lim x 2  0
x 0
x 0
1
e 
e x  ln(1  x)  1
1 x
lim

lim
x 0
x 0
x2
2x
1
x
e 
(1  x) 2
 lim
1
x 0
2
x
20
The Indeterminate Forms of
Type  
If
g ( x)  
lim f ( x)   and lim
x u
x u
Then
lim
x u
21
f ( x)
f ' ( x)
 lim
g ( x)
x u g ' ( x )
Example (∞/∞)

x  2 x2
Find lim
x  3 x 2  5 x
lim( x  2 x 2 )  , lim  3x 2  5 x   
x 0
x 0
x  2 x2
1  2x
lim 2
 lim
x  3 x  5 x
x  6 x  5
2
1
 lim

x  6
3
22
p
x
Example: Find lim x , where p>0。
x e
x
p
lim x  , lim e  
x 
x
xp
px p 1
 lim x  lim x
x  e
x  e
lim px ( p 1)    1  p  k  0
x 
k N
xp
px p 1
lim x  lim x
x  e
x  e

p( p  1) ( p  k  1) x p k
 lim
0
x
x 
e
23
Example: Find
sec x
lim 
x (  / 2 ) 1  tan x
(∞/∞)
sec x
sec x tan x
lim 
 lim 
x ( / 2 ) 1  tan x
x (  / 2 )
sec 2 x
 lim  sin x  1
x ( / 2 )
24
Example: Find
ln x
lim
x 2 x
(∞/∞)
ln x
1/ x
lim a  lim
x  x
x  1 /
x
1
 lim
0
x 
x
25
ln x
Example: Find lim
(a>0) (∞/∞)
x  x a
lim ln x   lim x a  
x 
x
ln x
1/ x
1
lim a  lim a 1  lim a  0
x  x
x  ax
x   ax
26
(ln x ) 2
Example: Find lim x (∞/∞)
x 
2
lim(ln x )2   lim 2 x  
x 
x
1
( 2 ln x ) 
2
( 2 ln x )
(ln x )
x

lim
lim

lim
x
x
x  (ln 2)  x  2 x
x 
x


2
(ln 2)  2
2/ x
0
 lim
x
x
x  (ln 2)  [2  (ln 2)  x  2 ]
27
ln x
Example: Find xlim
0 cot x
lim ln x  
x 
(∞/∞)
lim cot x  
x 0
2
ln x
1/ x

sin
x
lim
 lim
 lim
2
x 0 cot x
x 0  csc x
x 0
x

sin x
  lim sin x lim
x 0
x 0
x
 0  1  0
28
The Indeterminate Forms
0   and   
f  x  g  x  0  
To evaluate lim
n z
Rewrite
Or
f  x
0
f  x  g  x 

1 g  x 0
g  x

f  x  g  x 

1 f  x 
Then apply L’Hospital’s Rule
29
The Indeterminate Forms
0   and   
 f  x   g  x     
To evaluate lim
n z 
F(x)-g(x) must rewrite as a single
term. When the trigonometric
functions are involved, switching to
all sines and cosins may help.
30
1/ 2
x ln x
Example: Find xlim
0

lim x1 / 2  0
x 0
1/ 2
lim x
x 0
31
lim ln x  
x 0
 0
ln x   
ln x  lim 1/ 2  
x 0 x

1/ x
 lim
 lim  2 x  0
1 3 / 2 x0
x 0
 x
2
1
Example: Find lim x sin
 0
x
x
1
sin
1
x
lim x sin  lim
x 
x x  1
x
sin t
 lim
t 0
t 1
32
lim
(tanx  ln sin x )  0
Example: Find x

( )

2
lim
tan x  
 
x ( 2 )
lim
ln sin x  0
 
x( 2 )
ln sin x  0 
lim
(tan
x

ln
sin
x
)

lim
 

x ( 2 )
x ( 2 ) cot x
0
1
 cos x
sin x
 lim
(  cos x  sin x )  0
 lim


x( 2 )
x ( 2 ) 
 csc2 x
33
x
1
Example: Find lim (

) (∞−∞)
x 1 x  1
ln x
x
lim

x 1 x  1
1
lim

x 1 ln x
x
1
x ln x  x  1
lim(

)  lim

x 1 x  1
ln x x1 ( x  1) ln x
 00 
x ln x
ln x  x  1 / x  1  lim
 lim
x 1 x ln x  x  1
x 1 ln x  ( x  1)  1 / x

ln x  1 1
 lim

x 1 ln x  2
2
34
1
1
(

) (∞−∞)
Example: Find lim
x 1 ln x
x 1
1
1
x  1  ln x  0 
lim(

)  lim
 
x 1 ln x
x

1
x 1
( x  1) ln x  0 
1  1/ x
x 1
 lim
x 1
x  1  lim
x 1 x ln x  x  1
ln x 
x
1
1
 lim

x 1 2  ln x
2
35
Example: Find lim
(sec x  tan x ) (∞−∞)

x ( 2 ) 
lim
sec x  
 
x ( 2 )
lim
tan x  
 
x ( 2 )
1  sin x
lim
(sec x  tan x)  lim


x ( 2 )
x ( 2 ) cos x
 cos x
 lim
0


x ( 2 )  sin x
36
 00 
[ln 2 x  ln( x  1)]
Example: Find lim
x 
lim ln 2 x  
x 
lim ln( x  1)  
x 
2x
lim[ln 2 x  ln( x  1)]  lim ln
x 
x 
x 1
2x
 ln( lim
)  ln( lim 2 )  ln 2
x  x  1
x  1
37
  
1 0
Example: Find lim x tan  
x 
x
1
tan
1
x 0
lim x tan  lim
0
x 
x

1
x
x
2
sec
t
tan t
 lim
1
 lim
t 0
t 0
1
t
38

Example: Find lim(  x ) tan x  0
x

lim (

x 2
2
2
2
 x ) tan x  lim (
x 2

2
 x ) tan[( x 

2
)
  lim( t )(  cot t )
 lim( t ) tan( t  )
t 0
2
t 0
t cos t
t cos t
t
 lim
 lim
 lim
lim cos t
t 0 sin t
t 0 sin t
t 0 sin t t 0
1
 lim
lim cos t  1  1  1
t 0 cos t t 0
39

2
]
The Indeterminate Forms

0
0
0 ,  and 1
In these cases
g  x
y

f
x


1. Let
  
2. ln y  g  x  ln  f  x  
g  x  ln  f  x   exists and equal L,
3. If lim
x a
g  x
then lim  f  x   e L
x a
40
Example: Find
lim (1  x )  1
x 0
lim (1  x )
x 0
lim (1  x )cot x
x 0
lim cot  
x 0
cot x
 lim e
lim (1  x)cot x  1
x 0
cot x ln(1 x )
x 0
ln(1  x )
 exp[ lim
]
x 0
tan x
1
 exp[lim
]1
2
x 0 (`1  x ) sec x
41
n
lim
Example: Find n n
let y  n
n
ln n
ln y 
n
1
ln x
lim ln y  lim
 lim x  0
n
x  x
n  1
and
lim ln y  ln( lim y )  0  lim y  1
n 
Then
42
n 
lim n n  1
n 
n 
1 x
(1  )
Example: Find lim
x 
x
1 
1
lim(1  )  1 lim x  
x 
x 
x

1
x ln(1 )
1 x
x
lim(1  )  lim e
x 
x 
x
1
 exp[ lim x ln(1  )]
x 
x
1
lim x ln(1  )  0   
x 
x
43
1
ln(1  )
1
x
 lim x ln(1  )  lim
x 
1
x x
x
ln(1  t )
 lim
t 0
t

Then
44
1 x
lim(1  )  exp( 1)  e
x 
x
 00 
1
 lim 1  t  1
t 0
1
cot x
lim
(
1

sin
4
x
)
Example: x0
。
lim (1  x )  1 lim cot  
x 0

1
x 0
lim (1  x )cot x  lim ecot x ln(1sin 4 x )
x 0
x 0
ln(1  sin 4 x )
 exp[ lim
]
x 0
tan x
4 cos4 x
4
 exp[lim
]

e
2
x 0 (`1  sin 4 x ) sec x
45
x
0

lim
x
Example: Find

0
x 0

x ln x )
lim x x  lim e x ln x  exp( xlim
0
x 0
x 0
1
ln x
 lim x  lim (  x )  0
lim x ln x  lim
1 x 0
x 0
x0
x0 1
 2
x
x

 lim x  exp( lim x ln x)  exp(0)  1
x
x 0
46
x 0
1 x 0
Example: Find lim ( 2 )   
x 0 x
1 x
lim( 2 )  lim e
x 0 x
x 0
x ln
1
x2
1
 exp( lim x ln 2 )
x0
x
1
x ln 2
Replace the result of lim
x 0
x
47
1
ln 2
1
lim x ln 2  lim x
x 0
x 0
1
x
x
2 2
x  3
x
 lim
x 0
1
 2
x
 lim 2 x  0
x 0
1 x
1
Then lim( 2 )  exp( lim x ln 2 )  exp( 0)  1
x 0 x
x 0
x
48
Example: lim(e  x )
x
1/ x
x 0
lim(e  x)
x
x 0
1/ x

1
 
1 / x ln( e x  x )
 lim e
x0
ln(e  x )
 lim exp(
)
x 0
x
ln(e x  x )
 exp(lim
)
x 0
x
x
49
(e  x )
Example: lim
x 0
x
1/ x
e 1
and
x
x
ln(e  x )
lim
 lim e  x  1
x 0
x 0
x
1
x
Then lim(e x  x )1/ x
x 0
ln(e x  x )
 exp(lim
)
x 0
x
 exp(1)  e
50