Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
11.4 The l’Hospital’s rule (4.5)
Cauchy’s MVT: Let f, g be continuous on [a, b] and differentiable on ( a, b ) , and
assume g ' ( x ) ≠ 0 . There is c ∈ ( a, b ) such that
f ' ( c ) f (b ) − f ( a )
=
g ' ( c ) g (b ) − g ( a )
It is easy to see
f ' ( c ) f (b ) − f ( a )
f (b ) − f ( a )
b−a
, but why the same c for
=
=
g ' (c )
b−a
g (b ) − g ( a ) g (b ) − g ( a )
both?
Define h ( x ) = f ( x ) + rg ( x ) , where r is used to define h such that h ( a ) = h (b ) , thus
h ( a ) = f ( a ) + rg ( a ) = f (b ) + rg (b ) = h (b )
⇒ f ( a ) − f (b ) = r ( g (b ) − g ( a )) ⇒ r = −
f (b ) − f ( a )
g (b ) − g ( a )
Now, by Rolle’s theorem there is
0 = h ' (c ) = f ' (c ) −
Ex 10.
f (b ) − f ( a )
f ' ( c ) f (b ) − f ( a )
g ' (c ) ⇒
=
g (b ) − g ( a )
g ' ( c ) g (b ) − g ( a )
Show that 1 − cos x < x2 / 2, x ≠ 0 è
cos x
sin c
=
< 1 for some c ∈ ( 0, x )
2
1− x / 2
c
The L’Hospital’s Rule: The good use of Cauchy MVT is in the L’Hospital’s
(pronounced Loopeetal) rule, which say if lim
x→a
lim
x→a
f ( x)
"0" " ∞ "
is of the
form then
,
g ( x)
"0" " ∞ "
f ( x)
f '( x)
.
= lim
g ( x ) x →a g ' ( x )
If f(x), g(x) aren’t defined at x=a, the discontinuity is removable. Therefore we can
consider f(a)=0=f(b).
lim
x→a
Ex 1.
f ( x)
f ( x) − 0
f ( x) − f (a )
f ( x) − f (a )
f ' (c )
= lim
= lim
= lim
= lim
g ( x ) x →a g ( x ) − 0 x →a g ( x ) − f ( a ) x →a g ( x ) − g ( a ) c →a g ' ( c )
sin x
cos x
= lim
=1
x →0
x →0
x
1
lim
Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
sin h − h
cos h − 1
− sin h
= lim
= lim
=0
2
h →0
h →0
h →0
h
2h
2
Ex 2.
lim
Ex 3.
lim x ln x = lim
Ex 4.
lim x x = lim eln x = lim e x ln x = e0 = 1
Ex 5.
eax − 1
aeax
= lim
=a
x →0
x →0 1
x
Ex 6.
lim
ln x
1/ x
x2
= lim
=
lim
−
= lim x = 0
x →0 1/ x
x →0 −1/ x 2
x →0
x x →0
x →0
x
x →0
x →0
x →0
lim
e x − e− x
e x + e− x
2
2e
= lim
= −1
=
−
1
x →0 ln ( e − x ) + x − 1
x →0
− ( e − x ) + 1 −e + 1 e − 1
1 + x − (1 + 2x − x8 )
2
1
2 1+ x
− 12 + 4x
Ex 7.
lim
Ex 8.
1 ! 1 $ 2
1
1
&& x cos
2x
sin
+ ## −
x sin
2
x " x %
x
x = ∃ lim
lim
x→0 sin x
x→0
cos x
1
x 2 sin
x = lim x lim x sin 1 = 1⋅ 0 = 0
but: lim
x→0 sin x
x→0 sin x x→0
x
x
x →0
2
3
= lim
x →0
3x
2
−
3
3
− 1 (1 + x ) 2 + 14
(1 + x )
= lim 4
= lim 8
x →0
x →0
6x
6
− 52
=
1
16