L'Hôpital's rule Theorem
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L20
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L'Hôpital's rule Theorem Let
f , g : (a,b)
R be two functions of class C n , with n
³ 1, on the open interval (a,b). Let x 0 be a
point of (a,b) such that (i)
f (x 0 ) = 0, f HkL (x 0 ) = 0, k = 1, ... , n-1; (ii) g (x 0 ) = 0,
g HkL (x 0
(iii) g HnL (x 0 ) ¹ 0.
Then Lim x®x 0 fHxL gHxL =
f HnL Hx 0 L g HnL Hx 0 L . Proof Since
g HnL (x 0
¹ 0 and
g HnL is a contiuous function, we may assume without loss of generality that g HnL (x) ¹ 0, for every x. Then, by Taylor's theorem, there is some open interval (x 0 - ∆ ,x 0 + ∆ ) Ì (a,b), such that f (x)=
f (x 0 ) + f '(x 0 )(x - x
0 ) +
1 2 f ''(x 0 )(x - x 0 ) 2 + ... + 1 Hn- 1 L! f Hn- 1 L
0 )(x - x
0 )
1 +O
(x - x 0 ) = O n (x - x
0 ) where O n (x - x
0 ) is a continuous function such that Lim
0
n Hx-x 0 L
0 L
= f
HnL (x 0 ) and
g (x) =
g (x 0 ) + g '(x 0 )(x - x
0 ) +
1 2 g ''(x 0 )(x - x 0 ) 2 + ... + 1 Hn- 1 L! g Hn- 1 L
0 )(x - x
0 )
1 + O'
n (x -
x 0 ) = O' n (x - x
0 ) where O' n (x - x
0 ) is a continuous function such that Lim
0
'
Hx-x 0 L
0 L
= g
HnL (x 0 ) ¹ 0. Lim x®x 0
'
Hx-x 0 L
0 L
= g
HnL (x 0 ) ¹ 0. Then, Lim
x®x 0 fHxL gHxL = = Lim x®x 0
n Hx-x 0 L
' n Hx-x 0 L
x®x 0 J O n Hx-x 0 L
0 L
· Hx-x 0 L
O '
Hx-x 0 L N = f HnL Hx 0 L
HnL Hx 0 L
à
Evaluate the limit Lim
x® 0 Exp Ix 2 M - 1 Sin
Ix 2 M .
f (x) = Exp(x 2 ) - 1 and g (x) = Sin(x 2 )
1.0 - 0.5 0.5 1.0
1.2 1.4
1.6 1.8
2.0 à Then g ' (x) = 2xCos(x 2 ), g ''(x) = 2Cos(x 2 ) - 4x
2 Sin(x
2 ) Þ g ''(0) = 2 ¹ 0.
2
and
f ' (x) = 2 ã x 2 x,
f ''(x) = 2 ã x
+ 4 ã x 2 x 2
Þ f ''(0) = 2 Thus Lim
x® 0 Exp Ix 2 M - 1 Sin
Ix 2 M = 1. Example Evaluate the limit Lim
0 Cos HxL - 1
2 .
(x) = Cos(x)-1 and g (x) = x 2 . - 20 - 10 10 20 - 0.5 - 0.4 - 0.3
- 0.2
- 0.1
à Then
g ' (x) = 2x, g ''(x) = 2 Þ
g ''(0) = 2 ¹ 0 L20.nb
and
f ' (x) = -Sin(x), f ''(x) = -Cos(x) Þ
f ''(0) = -1 . Therefore Lim
0 Cos HxL - 1
2 . = -1/2
Then Cos(x) - 1 » (1/2)x
2 as x
® 0.
What about the possible asymptotes ? Lim
x®¥ ( Cos HxL - 1
2 .)/x = Lim x®¥
Cos HxL - 1
. = 0 & Lim
( Cos HxL - 1
2 ) = 0 .
then the horizontal axis is an horizontal asymptote for x ®¥ . Example Evaluate the limit 3) Lim
1 Hx - 1 M 3 Log HxL . 0.5
1.0 1.5
2.0 2.5
3.0 1 2 3 4 5 6 7
à
f (x) = (x -1) 3 and g (x) = Log(x) Then
' (x) = 1/x Þ
g ''(1) = 1 ¹ 0.
and f ' (x) = 3(x -1) 2
Þ f ''(1) = 0. Therefore Lim
x® 1 Hx - 1 M 3 Log HxL = 0.
Let
f : I
R be a differentiable function defined on the open interval I. i) If
f is increasing (resp. decreasing) on I, then f '(x)
³ 0, for every x Î I (resp. f '(x)
b 0, for every x Î I)
f '(x)
³ 0 (resp. f '(x)
b 0), for every x Î I then f is increasing (resp. decreasing) on I. iii) If f '(x)>0 (resp. f '(x)<0), for every x Î I then f is strictly increasing (resp. strictly decreasing) on I. iv) If
f '(x
0 ) = 0 and f ''(x
0 ) > 0 then x 0 is a local minimum. v) If f '(x 0 ) = 0 and f ''(x
0 ) < 0 then x 0 is a local maximum. vi) If f '(x 0 ) = 0,
f ''(x
0 ) = 0 and f '''(x
0 ) ¹ 0 then x 0 is a horizontal inflection point. Definition L20.nb
Let f : I
R be a differentiable function defined on the open interval I and let x 0 Î I. We denote by
G f = {(x,
f (x)) : x Î I}
Ì
R 2 the graph of f and by T( G f )| x 0 the tangent line to G f at the point (x 0 , f (x 0 )), defined by the equation y = f (x 0 ) +
f '(x
0 )(x - x
0 )
Example f @ x_ D : = H x - 2 L H x + 3 L H x - 5 L ; a : = - 5; b : = 8; h1 : = 10; h2 : = 40; h : = 120; AR : = 1 2; Program Fig @ 5 D - 4 - 2 2 4 6 8 - 150
- 100
- 50 50 100 150
200 à
Given h ¹ 0 such that x 0 +h
Î I, we consider the line L(h) passingh through the points (x 0 ,
(x 0 )) and (x 0 +h, f (x 0 +h)). When h ® 0, then L(h) tends to the tangent line T( G f )| x 0 . 6
f @ x_ D : = H x - 2 L H x + 3 L H x - 5 L ; a : = - 5; b : = 8; h1 : = 10; h2 : = 40; h : = 120; AR : = 1 2; Program Fig @ 4, 2 D - 4 - 2 2 4 6 8 - 150
- 100
- 50 50 100 150
200 Fig @ 4, 1 D - 4 - 2 2 4 6 8 - 150
- 100
- 50 50 100 150
200 L20.nb
Fig @ 4, 0.001 D - 4 - 2 2 4 6 8 - 150
- 100
- 50 50 100 150
200 à
Let f
R be a differentiable function defined on the open interval I and let x 0 Î I. The half space strictly above the tangent line T( G f
x 0 is denoted by H + ( f ,x 0 ) and similarly the half space strictly below the tangent line T( G f )| x 0 is denoted by H - ( f ,x 0 ). They are defined by the inequalities H + ( f ,x 0 ) = {(x,y) Î
R 2 : y - f (x 0 ) - f '(x 0 )(x - x
0 ) > 0}
and H - ( f ,x 0 ) = {(x,y) Î
2 : y -
f (x 0 ) - f '(x 0 )(x - x
0 ) < 0}
Example f @ x_ D : = H x - 2 L H x + 3 L H x - 5 L ; a : = - 5; b : = 8; h1 : = 10; h2 : = 40; h : = 120; AR : = 1 2; Program 8
Fig @- 4 D - 4 - 2 2 4 6 8 - 150
- 100
- 50 50 100 150
200 à
Given a function f : I
R , a point x 0 Î I and an open neighbourhooh (x 0 - Ε , x
0 + Ε ) Ì I we denote by G f (x 0 , Ε ) the portion of the graph G f given by
G f (x 0 , Ε ) = {(x, f (x)) : x Î (x
0 - Ε , x 0 + Ε )}
Ì
G f . The function f : I
R is convex at x
0 Î I if there is a neighbourhood (x 0 - Ε , x 0 + Ε ) Ì I such that
G f (x 0 , Ε ) = {(x, f (x)) : x Î (x
0 - Ε , x 0 + Ε )}
Ì H + ( f ,x 0 ). We say that f is
stricly convex if (x,
f (x))
Î H + ( f ,x 0 ), for every x Î (x
0 - Ε , x 0 + Ε ) such that x ¹ x
. Example f @ x_ D : = H x - 2 L H x + 3 L H x - 5 L ; a : = - 5; b : = 8; h1 : = 10; h2 : = 40; h : = 120; AR : = 1 2; Program L20.nb
Fig @ 3, 3 D - 4 - 2 2 4 6 8 - 150
- 100
- 50 50 100 150
200 à
The function f : I
R is concave at x
0 Î I if there is a neighbourhood (x 0 - Ε , x 0 + Ε ) Ì I such that
G f (x 0 , Ε ) = {(x, f (x)) : x Î (x
0 - Ε , x 0 + Ε )}
Ì H - ( f ,x 0 ). We say that f is
stricly concave if (x,
f (x))
Î H - ( f ,x 0 ), for every x Î (x
0 - Ε , x 0 + Ε ) such that x ¹ x
. Definition Given a function f : I
R we say that x 0 Î I is an inflection point (or flex point) if there is an open neighbourhood (x 0 -
, x 0 + Ε ) Ì I such that either (i) (x, f (x))
Î H - ( f ,x 0 ),
" x Î (x 0 - Ε , x
0 ] and (x, f (x))
Î H + ( f ,x 0 ),
" x
Î [x
0 , x
0 + Ε ) or else
(ii) (x, f (x)) Î H + ( f ,x 0 ),
" x Î (x 0 - Ε , x
0 ] and (x, f (x))
Î H - ( f ,x 0 ),
" x
Î [x
0 , x
0 + Ε ) In the first case x 0 is an
ascending inflection point and in the second case x 0 is a
descending inflection point .
10
f @ x_ D : = - H x - 2 L H x + 3 L H x - 5 L ; a : = - 5; b : = 8; h1 : = 10; h2 : = 40; h : = 120; AR : = 1 2; Program Fig @ 4 3, 4 3 D - 4 - 2 2 4 6 8 - 200
- 150
- 100
- 50 50 100 150
à à
If f
R is function having first and second order derivatives on I and let x 0
Î I be a point of the domain of definition, then : i) if
f is convex at x 0
there is an open neighbourhood (x 0 - ¶ , x
0 + ¶ ) Í I such that f ''(x)
³ 0 for
every x Î (x 0 - ¶ , x 0 + ¶ ); ii) if there is an open neighbourhood (x 0 - ¶ , x 0 + ¶ ) Í I such that f ''(x) ³ 0 for every x Î (x
0 - ¶ , x 0 + ¶ )
f
f is convex at x 0 ; iii) if there is an open neighbourhood (x 0 - ¶ , x
0 + ¶ ) Í I such that f ''(x) > 0 for every x Î (x
0 - ¶ , x 0 + ¶ )
f
f is strictly convex at x 0 ; Theorem L20.nb
If f : I
R is function having first and second order derivatives on I and let x 0
Î I be a point of the domain of definition, then : i) if is f concave at x 0
there is an open neighbourhood (x 0 - ¶ , x
0 + ¶ ) Í I such that f ''(x)
£ 0 for
every x Î (x 0 - ¶ , x 0 + ¶ ); ii) if there is an open neighbourhood (x 0 - ¶ , x 0 + ¶ ) Í I such that f ''(x) £ 0 for every x Î (x
0 - ¶ , x 0 + ¶ )
f
f is concave at x 0 ; iii) if there is an open neighbourhood (x 0 - ¶ , x
0 + ¶ ) Í I such that f ''(x) < 0 for every x Î (x
0 - ¶ , x 0 + ¶ )
f
f is strictly concave at x 0 ; Theorem If
f : I
R is function having first and second order derivatives on I and let x 0
Î I be a point of the domain of definition, then : i) if x
0 is an inflection point
''(x 0 ) = 0; ii) if there is an open neighbourhood (x 0 - ¶ , x
0 + ¶ ) Í I such that f ''(x)
£ 0 for every x Î (x
0 - ¶ , x 0 ) and f ''(x) ³ 0 for every x Î (x
0 - ¶ , x 0 ) then x 0 is an ascending flex point. iii) if there is an open neighbourhood (x 0 - ¶ , x
0 + ¶ ) Í I such that f ''(x)
³ 0 for every x Î (x
0 - ¶ , x 0 ) and f ''(x) £ 0 for every x Î (x
0 - ¶ , x 0 ) then x 0 is a descending flex point. Example Let consider the hyperbolic tangent Tanh(x) = Sinh
HxL Cosh
HxL =
ex -
-
+
-
=
e 2
1
2
1 -
- 5 5 10 - 1.0 - 0.5
0.5 1.0
12
Problem 1 : show that Tanh(x) is strictly increasing. DTanh(x) = DSinh
HxL Cosh
HxL -
Sinh HxL DCosh HxL Cosh HxL 2 =
Cosh HxL 2 -
HxL 2 Cosh HxL 2 = 1 Cosh
Hx M 2 > 0. Problem 2 : find the asymptotes of Tanh(x) when x ® ±
They are the horizontal lines y = 1 and y = -1 respectively (Why?) so Im(Tanh(x)) = (-1,1). Need to compute
m = Lim x®±¥
fHxL x = 0, q = Lim x®±¥ ( f (x) - mx) = ±1 Problem 3 : find Tanh - 1
y = e 2
- 1
2 x + 1 Þ ( e 2
+ 1)y =
e 2
- 1
e 2
(y - 1) = - (1+y)
e 2 x =
1 +
1 -
x = 1 2 Log K 1 + y 1 - y O
x = Tanh - 1 (y) = 1 2 Log K 1 + y 1 - y O
: show that Tanh(x) is strictly convex for x < 0, strictly concave for x > 0 and x = 0 is a descending inflection point. D 2
1 Cosh
Hx M 2 ) = - 2 Sinh
HxL Cosh
HxL 3 Since Cosh(x) > 0, for every x and Sinh(x) = :
0 if
0, 0 if x = 0, > 0 if x > 0 then D 2 Tanh(x) = : > 0 if x < 0, 0 if x = 0,
0 if
0.
D 2 Tanh(x) = : > 0 if x < 0, 0 if x = 0,
0 if
0. using theorem 18.2.7 we get the result. 14
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