8.1 L’Hôpital’s Rule
Actually, L’Hôpital’s Rule was
developed by his teacher
Johann Bernoulli. De
L’Hôpital paid Bernoulli for
private lessons, and then
published the first Calculus
book based on those lessons.
Guillaume De l'Hôpital
1661 - 1704
8.1 L’Hôpital’s Rule
Consider:
x2  4
lim
x2 x  2
If we try to evaluate this by
direct substitution, we get:
Zero divided by zero can not be evaluated, and is an example
of indeterminate form.
In this case, we can evaluate this limit by factoring and
canceling:
x2  4
lim
x2 x  2
x  2  x  2 

 lim
x 2
x2
 lim  x  2   4
x 2
0
0
8.1 L’Hôpital’s Rule
f  x
lim
xa g  x 
x2  4
 lim
x 2 x  2
x2  4
The limit is the ratio of
the numerator over the
denominator as x approaches 2.
4
0.05
3
2
1
-3
-2
-1
1
0
-1
-2
x
2
3
If we zoom in far enough,
the curves
will2 appear as2.05
0 1.95
x
straight lines.
-3
x2
-4
-0.05
-5
8.1 L’Hôpital’s Rule
0.05
f  x
lim
xa g  x 
x2  4
 lim
x 2 x  2
As
f  x
g  x
df
x 2
becomes:
dg
0 1.95
-0.05
2 dx
x
2.05
df
df  dx
dg dg
dx
8.1 L’Hôpital’s Rule
f  x
lim
xa g  x 
d 2
x  4

x2  4
dx
 lim

lim
x 2 x  2
x 2 d
 x  2
dx
2x
 lim
4
x2 1
8.1 L’Hôpital’s Rule
L’Hôpital’s Rule:
f  x
If lim
is indeterminate, then:
xa g  x 
f  x
lim
 lim
x a g  x 
x a
f  x
g x
8.1 L’Hôpital’s Rule
We can confirm L’Hôpital’s rule by working backwards, and
using the definition of derivative:
f a
g  a 
f  x  f a
lim
xa
xa

g  x  g a
lim
xa
xa
f  x  f a
 lim
x a g  x   g  a 
f  x  f a
xa
 lim
xa g  x   g  a 
xa
f  x  0
f  x
 lim
 lim
xa g  x   0
x a g  x 
8.1 L’Hôpital’s Rule
Example:
1  cos x  lim sin x
0
lim
x

0
2
1

2
x
x 0 x  x
If it’s no longer
indeterminate, then STOP!
If we try to continue with L’Hôpital’s rule:
sin x
x 0 1  2 x
 lim
cos x
1

x 0
2
2
 lim
which is wrong,
wrong, wrong!
8.1 L’Hôpital’s Rule
On the other hand, you can apply L’Hôpital’s rule as many
times as necessary as long as the fraction is still
indeterminate:
x
1 x 1
2
lim
x 0
x2
1
2
1
1  x   1  x
2
lim
x 0
x2
0
0
1
1
1

2
1  x  
2
 lim 2
x 0
2x
3
1

 1  x  2
not
 lim 4
x 0
2
1

 4
2
1

8
0
0
0
0
8.1 L’Hôpital’s Rule
L’Hôpital’s rule can be used to evaluate other indeterminate
forms besides


0
.
0
The following are also considered indeterminate:
 0


1
0

The first one,
, can be evaluated just like

The others must be changed to fractions first.
0
0
.
0
0
8.1 L’Hôpital’s Rule
1

This approaches
lim  x sin 

0
x 
x

1
sin
x
0
lim
This approaches
x 
1
0
x
1  1 
1
cos      2 
sin
x   x   lim cos  1 

x
 cos  0  1
lim
 lim


x

x 
x 
1
1
x
 2
x
x
8.1 L’Hôpital’s Rule
Indeterminate Forms:
lim f  x   L

x a
1/ x

lim x
x 

lim ln x1 / x
e x

0
e
e
1
0
00
lim f  x   lim e
x a
x a
1
lim ln  x 
x x
e
ln  x 
lim
x x


1
lim x
x 1
e
0
ln f  x 
 eL
L’Hôpital
applied
1
8.2 Relative Rates of Growth
x
y

e
The function
grows very fast.
If x is 3 inches,
y is about 20
inches:
35
30
25
20
15
10
5
3, 20
0
20
We
have
gonethe
lessy-value
than halfAt 64
inches,
way across the board
would be at the edge of the
horizontally,
and already the yknown universe!
value
the
(10.5 would
billion reach
light-years)
Andromeda Galaxy!
40
60
At x  10 inches,
At x  44 inches,
1
y  mile y  2 million light-years
3
80
8.2 Relative Rates of Growth
The function
y = ln x grows
very slowly.
We would have to
move 2.6 miles to the
right before the line
moves a foot above
the x-axis!
35
30
25
20
15
10
5
0
By the time we reach the edge
of the universe again (10.5
billion light-years) the chalk
line will only have reached 64
inches!
20
40
60
80
The function y = ln x increases everywhere, even though it
increases extremely slowly.
8.2 Relative Rates of Growth
Let f (x) and g(x) be positive for x sufficiently large.
Definitions: Faster, Slower, Same-rate Growth as
x 
1. f grows faster than g (and g grows slower than f )
as x  if
lim
x 
f  x
g  x

or
lim
x 
g  x
f  x
2. f and g grow at the same rate as
lim
x 
f  x
g  x
L0
0
if 
x
8.2 Relative Rates of Growth
WARNING
Please temporarily suspend your common sense.
8.2 Relative Rates of Growth
According to this definition, y = 2x does not grow faster than
yx
2x
lim
 lim 2  2
6
x  x
x 
5
4
3
2
1
-6 -5 -4 -3 -2 -1 0
-1
-2
-3
-4
-5
-6
yx
y  2x
Since this is a finite non-zero
limit, the functions grow at the
same rate!
1 2 3 4 5 6
The book says that “f grows faster
than g” means that for large x
values, g is negligible compared to f.
8.2 Relative Rates of Growth
Which grows faster,

ex
lim 2 
x  x

x
2
eor
x?
Weiscan
confirm thisso
graphically:
This
indeterminate,
we
10
apply L’Hôpital’s
rule.
9

e

lim
x  2 x

x
ex
y 2
x
8
7
Still indeterminate.
6
5
ex

lim
x  2
4
3
2
1
e
x
grows faster than
2
x.
0
1
2
3
4
5
6
7
8
9 10
8.2 Relative Rates of Growth
“Growing at the same rate” is transitive.
In other words, if two functions grow at the same rate as a
third function, then the first two functions grow at the same
rate.
8.2 Relative Rates of Growth
Show that f  x   x  5 and g  x    2 x  1 grow
2
at the same rate as x  .
2
8.2 Relative Rates of Growth
Let h  x   x
x  5  lim
lim
x 
x 
x
2

lim
x 

2 x 1
x
2
x2  5

 lim
x 
x2
x2 5
 lim 2  2
x 
x
x

2 x 1
 x
2
2
f
 f h
1 1
lim
 lim     1  
x  g
x  h g
4 4


50
 lim 1  2  1
x 
x
2
2 x
1 0
 lim 

  4
x 
x
 x
f and g grow at the
same rate.
8.2 Relative Rates of Growth
Definition f of Smaller Order than g
Let f and g be positive for x sufficiently large. Then f
is of smaller order than g as x   if
lim
x 
We write
f  x
g  x
0
f  o  g  and say “f is little-oh of g.”
Saying f  o  g  is another way to say that f grows
slower than g.
8.2 Relative Rates of Growth
Definition f of at Most the Order of g
Let f and g be positive for x sufficiently large. Then f
is of at most the order of g as x   if there is a
positive integer M for which
f  x
g  x
 M for x sufficiently large
f  O  g  and say “f is big-oh of g.”
Saying f  O  g  is another way to say that f grows
no faster than g.
We write
8.3 Improper Integrals
Until now we have been finding integrals of continuous
functions over closed intervals.
Sometimes we can find integrals for functions where
the function or the limits are infinite. These are called
improper integrals.
8.3 Improper Integrals
4

1
0
1 x
dx
1 x
The function is
undefined at x = 1 .
Since x = 1 is an asymptote, the
function has no maximum.
We could define this integral as:
lim 
b 1
b
0
1 x
dx
1 x
3
Can we find
the area under
an infinitely
high curve?
2
1
0
1
(left hand limit)
We must approach the limit from
inside the interval.
8.3 Improper Integrals
lim 
b 1
b
0
1 x
dx
1 x
1 x 1 x
dx
1 x 1 x


1+x
1  x2
dx
1

1
sin 1 x   u 2 du
2

1
1  x2
dx  
x
1  x2
u  1  x2
du  2x dx
1
 du  x dx
2
dx
8.3 Improper Integrals

1
1  x2
x
dx  
1  x2

1
2
1
sin x   u du
21
1
sin 1 x  u 2
b
dx
u  1  x2
du  2x dx
1
 du  x dx
2
This integral converges
because it approaches a
solution.
lim sin 1 x  1  x 2
b 1
0

0
0
2
1
lim sin b  1  b 2  sin 1 0  1
b 1

 



2
1
8.3 Improper Integrals
4
1
lim 
dx
b 0 b x
dx
0 x
1
1
3
2
1
lim ln x b
b0
lim ln1  ln b
1
b 0
-1
1
lim ln
b 0
b

0
1
-1
This integral diverges.
8.3 Improper Integrals
4

dx
3
0
 x  1
  x 1
3

The function
approaches 
when x 1 .
2
3
2
3
0
b 1
b1
2

3
dx  lim   x  1
3
c 1
0
lim 3  x  1
2
dx
lim   x  1
b
3
1 b
3
0
c
 lim 3  x  1
c 1
1 3
3
c
2

3
1
dx
0
1
2
3
8.3 Improper Integrals
lim   x  1
b
b 1

2
3
dx  lim   x  1
3
c 1
0
lim 3  x  1
1 b
3
b1
c
 lim 3  x  1
0
0

c 1
2
3
dx
1 3
3
c
0




lim 3  b  1  3  1   lim 3  2  3  c  1 
b 1 
 c1 

1
3
1
3
1
3
1
3
3  33 2
8.3 Improper Integrals


1
dx
P
x
b P 1
1 P 1
lim

b   P  1
P  1
P0
(P is a constant.)


1
x P dx
b
lim  x
b  1
P
What happens here?
 P 1
b
If P  1 then
gets bigger
and bigger as b   , therefore
the integral diverges.
dx
b
1
 P 1
lim
x
b   P  1
1
If P  1 then b has a negative
exponent and b  P 1  0 ,
therefore the integral converges.
8.3 Improper Integrals


1
x
e dx
b
Converges
lim  e dx
x
lim e
b  1
lim  e   e
b
b 
x b
b 
1

lim 
b 
1
0
1 1
e
b

e
1

e
8.3 Improper Integrals
Does


1
e
 x2
dx
converge?
Compare:
1
e
x
2
to
1
ex
For x  1, e
for positive values of x.
x2
e
x

1
e
x2
1
 x
e
8.3 Improper Integrals
For x  1, e
x2
e
x

1
e
Since
1
e
x2
x2
1
 x
e
1
is always below
,
x
e
we say that it is “bounded above” by
1
x
e
1
1
Since x converges to a finite number, x 2 must also converge!
e
e
8.3 Improper Integrals
Direct Comparison Test:

Let f and g be continuous on a,   with 0  f  x   g  x 
for all
xa
, then:


1
 f  x  dx
a

2
 g  x  dx
a
 g  x  dx converges.
converges if
diverges if
a

 f  x  dx diverges.
a
8.3 Improper Integrals


1
sin 2 x
dx
2
x
The maximum value of sin x  1 so:
2
sin x 1
0
 2
2
x
x
on
on
1,  
1
sin 2 x
Since 2 converges,
converges.
2
x
x
8.3 Improper Integrals


1
1
x 2  0.1
x2  0.1  x
1
dx
for positive values of x, so:
1

on
2
x  0.1 x
1
Since
diverges,
x
1,  
1
x  0.1
2
diverges.
8.3 Improper Integrals
Does


1
If functions grow at the same
rate, then either they both
converge or both diverge.
dx
2 converge?
1 x
x   the “1” in the denominator becomes
1
insignificant, so we compare to 2 .
x
1
1
Since 2 converges,
2
2
1 x
x
x
2x
 lim
lim
1
As
x 
1
1  x2
x 
x2
 lim
x 
2x
1
1  x2
converges.
8.3 Improper Integrals


1
dx
1  x2
 lim 
b
b  1
1
b
dx
1  x2 
2

4
1
 lim tan x  lim tan b  tan 1
b 
1


2

1
b

 
4 2


4
4
As y  , x 

2
y  tan x
8.3 Improper Integrals


1
dx
1  x2
 lim 
b
b  1

1
dx
1  x2
1
 lim tan x
b 


 lim  x 2 dx
b  1
b
1
2


4
1
dx
2
x
b
 lim tan 1 b  tan 1 1
b



4
 lim  x 1
b 
b
1
0
1 
1
 lim     
b 
b  1
1
8.4 Partial Fractions
5x  3
 x 2  2 x  3 dx
5x  3
 x  3 x  1
This would be a lot easier if we could
re-write it as two separate terms.
A
B


x  3 x 1
Multiply by the common
denominator.
5x  3  A  x  1  B  x  3
5x  3  Ax  A  Bx  B  3
5x  Ax  Bx
5  A B
3  A  B  3
3  A  3B
Set like-terms equal to
each other.
Solve two equations with
two unknowns.
8.4 Partial Fractions
5x  3
 x 2  2 x  3 dx
A
B
5x  3


 x  3 x  1 x  3 x  1
5x  3  A  x  1  B  x  3
5x  3  Ax  A  Bx  B  3
5x  Ax  Bx
5  A B
3  A  B  3
3  A  3B
5  A B
3   A  3B
3  A  3B
8  4B
5  A 2
2B
3 A
3
2
 x  3  x  1 dx
3ln x  3  2ln x  1  C
Solve two equations with
Thisunknowns.
technique is called
two
Partial Fractions
8.4 Partial Fractions
Good News!
The AP Exam only requires non-repeating linear factors!
The more complicated methods of partial fractions are
good to know, and you might see them in college, but they
will not be on the AP exam or on my exam.
8.4 Partial Fractions
6x  7
 x  2
2
A
B


2
x  2  x  2
6 x  7  A  x  2  B
Repeated roots: we must use
two terms for partial
fractions.
7  12  B
6x  7  Ax  2 A  B
6x  Ax
7  2A  B
6 A
7  26  B
5  B
6
5

x  2  x  2 2
8.4 Partial Fractions
2 x3  4 x 2  x  3
x2  2 x  3
If the degree of the numerator is higher
than the degree of the denominator,
use long division first.
2x
x 2  2 x  3 2 x3  4 x 2  x  3
2 x3  4 x 2  6 x
5x  3
5x  3
2x  2
x  2x  3
5x  3
2x 
 x  3 x  1
3
2
 2x 

 x  3  x  1
(from example one)
8.4 Partial Fractions
x
2 x  4
2

 1  x  1
irreducible
quadratic
factor
2
Ax  B
C
D
 2


x  1 x  1  x  12
repeated root




2 x  4   Ax  B  x  1  C x 2  1  x  1  D x 2  1
2




2 x  4   Ax  B  x 2  2 x  1  C x 3  x 2  x  1  Dx 2  D
2 x  4  Ax3  2 Ax 2  Ax  Bx 2  2 Bx  B  Cx 3  Cx 2  Cx  C  Dx 2  D
8.4 Partial Fractions

dx
4  x2
These are in
the same
form.
22
2
4

x
a x

2sec  d
 2sec
2
 sec d
ln sec  tan   C
4 x
x
 C
2
2
2
ln
x
2a
4  x2
sec 
2
2sec  4  x
2
x
tan  
2
2 tan  x
2sec 2  d  dx
8.4 Partial Fractions

dx
4  x2
2sec2  d
 2sec
4  x2  x
C
2
ln
ln
4  x 2  x  ln 2  C
 sec d
ln sec  tan   C
ln
4  x2 x
 C
2
2
This is a constant.
ln
4  x2  x  C
8.4 Partial Fractions
a2  x2
a2  x2,
If the integral contains

we use the triangle at right.
If we need
a2  x,2 we
move a to the hypotenuse.
a

a2  x2
x
a
If we need
x2  a,2 we
move x to the hypotenuse.
x
x
x a
2

a
2
8.4 Partial Fractions

x 2 dx
3
9  x2

9  x2
9sin 2   3cos d

3cos
1  cos 2
9
d
2
9
1  cos 2 d

2
9
9 1
   sin 2  C
2
2 2
x
x sin   3
3sin  x
9  x2
cos 
3
3cos  9  x2
3cos d  dx
x
1 x
sin  
  sin
3
3
9 1 x 9 x 9  x 2
sin
  
C
2
3 2 3
3
8.4 Partial Fractions

3
x 2 dx
9 x
2
x

9  x2
9 1  x  x
sin   
9  x2  C
2
3 2
9
9 1
   sin 2  C
2
2 2
9 1 x 9
sin
  2sin  cos   C
2
3 4
9 1 x 9 x 9  x 2
sin
  
C
2
3 2 3
3
8.4 Partial Fractions


dx
We can get 2x  x 2 into the necessary form
by completing the square.
2x  x2
2x  x 2
 x  2x

2


 x2  2x  1  1
  x  1  1
2
1   x  1
2

sin   u
dx
1   x  1
2
cos d  du
8.4 Partial Fractions


dx
2x  x
du
1 u2
cos  d
 cos
 d
2

dx
1   x  1
2
1
Let u  x 1
du  dx
1

sin
u C
  C
 sin 1  x 1  C
u

1 u2
1 u2
 1 u2
cos 
1
sin   u cos d  du
8.4 Partial Fractions
dx
 4 x2  4 x  2
Complete the square:
dx
4 x2  4 x  1  1
  2 x  1
2
1
4 x2  4 x  2
 2 x  1
2
1
8.4 Partial Fractions
dx
 4 x2  4 x  2
dx
  2 x  1
2
1
1 du
2  u2 1
1 sec 2  d
2  sec 2 
Let u  2x  1
du  2 dx
1
du  dx
2
u2 1
u

1
tan  u
sec 2  d  du
sec  u 2  1
1
1
2
2
d    C
sec


u
1

2
2
1
1
1
tan 1  2 x  1  C

tan u  C 
2
2
8.4 Partial Fractions
Here are a couple of shortcuts that are result from
Trigonometric Substitution:
du
1
1 u
 u 2  a 2  a tan a  C

du
u
 sin
C
a
a2  u 2
1
These are on your list of
formulas. They are not
really new.