guillaume de l'hôpital 1661 - 1704 actually, l’hôpital’s rule was developed by his...
TRANSCRIPT
Guillaume De l'Hôpital1661 - 1704
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.
8.1 L’Hôpital’s Rule
Zero divided by zero can not be evaluated, and is an example of indeterminate form.
2
2
4lim
2x
x
x
Consider: If we try to evaluate this by direct substitution, we get:
0
0
In this case, we can evaluate this limit by factoring and canceling:
2
2
4lim
2x
x
x
2
2 2lim
2x
x x
x
2lim 2x
x
4
8.1 L’Hôpital’s Rule
If we zoom in far enough, the curves will appear as straight lines.
2
2
4lim
2x
x
x
The limit is the ratio of the numerator over the denominator as x approaches 2.
2 4x
2x
limx a
f x
g x
8.1 L’Hôpital’s Rule
2
2
4lim
2x
x
x
limx a
f x
g xAs 2x
f x
g xbecomes:
df
dg
df
dg
dx
dxdf
xdgd
8.1 L’Hôpital’s Rule
2
2
4lim
2x
x
x
limx a
f x
g x
2
2
4lim
2x
dx
dxd
xdx
2
2lim
1x
x
4
8.1 L’Hôpital’s Rule
L’Hôpital’s Rule:
If is indeterminate, then:
limx a
f x
g x
lim limx a x a
f x f x
g 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
lim
lim
x a
x a
f x f a
x ag x g a
x a
limx a
f x f a
x ag x g a
x a
limx a
f x f a
g x g a
0lim
0x a
f x
g x
limx a
f x
g x
8.1 L’Hôpital’s Rule
Example:
20
1 coslimx
x
x x
0
sinlim
1 2x
x
x
0
If it’s no longer indeterminate, then STOP!
If we try to continue with L’Hôpital’s rule:
0
sinlim
1 2x
x
x
0
coslim
2x
x
1
2 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:
20
1 12lim
x
xx
x
1
2
0
1 11
2 2lim2x
x
x
0
0
0
0
0
0not
1
2
20
11 1
2limx
x x
x
3
2
0
11
4lim2x
x
14
2
1
8
8.1 L’Hôpital’s Rule
L’Hôpital’s rule can be used to evaluate other indeterminate0
0forms besides .
The following are also considered indeterminate:
0 1 00 0
The first one, , can be evaluated just like .
0
0
The others must be changed to fractions first.
8.1 L’Hôpital’s Rule
1lim sinx
xx
This approaches0
0
1sin
lim1x
x
x
This approaches 0
2
2
1 1cos
lim1x
x x
x
1sin
lim1x
x
x
1lim cosx x
cos 0 1
8.1 L’Hôpital’s Rule
Indeterminate Forms: 1 00 0
lnlim lim lim f x L
x a x a x af x L f x e e
1/lim x
xx
1/lim ln x
xx
e
1lim lnx
xxe
lnlimx
x
xe
1
lim1x
x
e
0e 1
0
L’Hôpitalapplied
8.1 L’Hôpital’s Rule
The function grows very fast.
xy e
If x is 3 inches,
y is about 20 inches:
3,20
At 10 inches,
1 mile
3
x
y
At 44 inches,
2 million light-years
x
y
We have gone less than half-We have gone less than half-way across the board way across the board horizontally, and already the y-horizontally, and already the y-value would reach the value would reach the Andromeda Galaxy!Andromeda Galaxy!
At 64 inches, the y-value At 64 inches, the y-value would be at the edge of the would be at the edge of the known universe!known universe!(10.5 billion light-years)(10.5 billion light-years)
8.2 Relative Rates of Growth
The functiony = 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!
By the time we reach the edge By the time we reach the edge of the universe again (10.5 of the universe again (10.5 billion light-years) the chalk billion light-years) the chalk line will only have reached 64 line will only have reached 64 inches!inches!
The function y = ln x increases everywhere, even though it increases extremely slowly.
8.2 Relative Rates of Growth
Definitions: Faster, Slower, Same-rate Growth as x
Let f (x) and g(x) be positive for x sufficiently large.
1. f grows faster than g (and g grows slower than f ) as ifx
limx
f x
g x or
lim 0x
g x
f x
2. f and g grow at the same rate as ifx
lim 0x
f xL
g x
8.2 Relative Rates of Growth
WARNINGWARNINGWARNING
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
2y x
y x
2lim lim 2 2x x
x
x
Since this is a finite non-zero limit, the functions grow at the same rate!
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, or ?xe 2x
2lim
x
x
e
x
This is indeterminate, so we apply L’Hôpital’s rule.
lim2
x
x
e
x
lim2
x
x
e
Still indeterminate.
grows faster than .xe 2x
We can confirm this graphically:
2
xey
x
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 and grow
at the same rate as .
2 5f x x 2
2 1g x x
x
8.2 Relative Rates of Growth
Let h x x
2 5limx
x
x
2
2
5limx
x
x
2
2 2
5limx
x
x x 2
5lim 1x x
10
2
2 1limx
x
x
2
2
2 1limx
x
x
2
2 1limx
x
x x
limx
f
g
40
limx
f h
h g
11
4 1
4 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 ifx
lim 0x
f x
g x
We write and say “f is little-oh of g.” of g
Saying is another way to say that f grows slower than g.
of g
8.2 Relative Rates of Growth
Saying is another way to say that f grows no faster than g.
Of g
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 if there is a
positive integer M for whichx
f xM
g x
We write and say “f is big-oh of g.” Of g
for x sufficiently large
8.2 Relative Rates of Growth
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
1
0
1
1
xdx
x
The function is
undefined at x = 1 .
Since x = 1 is an asymptote, the function has no maximum.
Can we find the area under an infinitely high curve?
We could define this integral as:
01
1lim
1
b
b
xdx
x
(left hand limit)
We must approach the limit from inside the interval.
8.3 Improper Integrals
01
1lim
1
b
b
xdx
x
1
1
1
1
x
x
xdx
x
2
1+x
1dx
x
2 2
1 x
1 1dx dx
x x
21u x 2 du x dx
1
2du x dx
11 2
1sin
2x u du
8.3 Improper Integrals
2 2
1 x
1 1dx dx
x x
21u x
2 du x dx1
2
du x dx 1
1 21
sin2
x u du
11 2sin x u
1 2
1 0lim sin 1
b
bx x
1 2 1
1lim sin 1 sin 0 1b
b b
1
2
2
0 0
This integral converges because it approaches a solution.
8.3 Improper Integrals
1
0
dx
x1
0lim ln
bbx
0lim ln1 lnb
b
0
1lim lnb b
This integral diverges.
1
0
1lim
bbdx
x
8.3 Improper Integrals
3
2031
dx
x The function
approacheswhen .
1x
233
01 x dx
2 233 3
01 1lim 1 lim 1
b
cb cx dx x dx
31 1
3 31 1
0
lim 3 1 lim 3 1b
b cc
x x
8.3 Improper Integrals
2 233 3
01 1lim 1 lim 1
b
cb cx dx x dx
31 1
3 31 1
0
lim 3 1 lim 3 1b
b cc
x x
11 1 133 3 3
1 1lim 3 1 3 1 lim 3 2 3 1b c
b c
0 0
33 3 2
8.3 Improper Integrals
1 P
dx
x
0P
1 Px dx
1lim
b P
bx dx
1
1
1lim
1
bP
bx
P
1 11lim
1 1
P P
b
b
P P
What happens here?
If then gets bigger and bigger as , therefore the integral diverges.
1P 1Pb
b
If then b has a negative exponent and ,therefore the integral converges.
1P 1 0Pb
(P is a constant.)
8.3 Improper Integrals
1
xe dx
1lim
b x
be dx
1lim
bx
be
1lim b
be e
1 1
lim bb e e
0 1
e
Converges
8.3 Improper Integrals
Does converge?2
1
xe dx
Compare:
to for positive values of x.2
1xe
1xe
For2
2x
1 11,
ex x
xx e e
e
8.3 Improper Integrals
For2
2x
1 11,
ex x
xx e e
e
Since is always below ,
we say that it is “bounded above” by
2
1xe
1xe
1xe
Since converges to a finite number, must also converge!1
xe2
1xe
8.3 Improper Integrals
Direct Comparison Test:
Let f and g be continuous on with
for all , then:
,a 0 f x g x
x a
2 a
f x dx
diverges if diverges. a
g x dx
1 a
f x dx
converges if converges. a
g x dx
8.3 Improper Integrals
2
21
sin
xdx
x
The maximum value of so:sin 1x
2
2 2
sin 10 on 1,
x
x x on
Since converges, converges.2
1
x
2
2
sin x
x
8.3 Improper Integrals
21
1
0.1dx
x
for positive values of x, so:2 0.1x x
Since diverges, diverges.1
x 2
1
0.1x
2
1 1
0.1 xx
on 1,
8.3 Improper Integrals
If functions grow at the same rate, then either they both converge or both diverge.Does converge?21 1
dx
x
As the “1” in the denominator becomes
insignificant, so we compare to .
x
2
1
x
2
2
1
lim1
1
x
x
x
2
2
1limx
x
x
2
lim2x
x
x 1
Since converges,
converges.
2
1
x2
1
1 x
8.3 Improper Integrals
21 1
dx
x
21lim
1
b
b
dx
x
1
1lim tan
b
bx
1 1lim tan tan 1
bb
2 4
4
tany xAs ,
2y x
2
2
4
4
8.3 Improper Integrals
21 1
dx
x
21lim
1
b
b
dx
x
1
1lim tan
b
bx
1 1lim tan tan 1b
b
2 4
4
21
1 dx
x
2
1lim
b
bx dx
1
1lim
b
bx
1 1lim
1b b
10
8.3 Improper Integrals
2
5 3
2 3
xdx
x x
This would be a lot easier if we couldre-write it as two separate terms.
5 3
3 1
x
x x
3 1
A B
x x
Multiply by the common denominator.
5 3 1 3x A x B x
5 3 3x Ax A Bx B Set like-terms equal to each other.
5x Ax Bx 3 3A B
5 A B 3 3A B Solve two equations with two unknowns.
8.4 Partial Fractions
2
5 3
2 3
xdx
x x
5 3
3 1
x
x x
3 1
A B
x x
5 3 1 3x A x B x
5 3 3x Ax A Bx B 5x Ax Bx 3 3A B
5 A B 3 3A B Solve two equations with two unknowns.
5 A B 3 3A B 3 3A B
8 4B2 B
5 2A 3 A3 2
3 1
dxx x
3ln 3 2ln 1x x C
This technique is calledPartial 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
2
6 7
2
x
x
Repeated roots: we must use two terms for partial fractions.
22 2
A B
x x
6 7 2x A x B
6 7 2x Ax A B
6x Ax 7 2A B
6 A 7 2 6 B
7 12 B
5 B
2
6 5
2 2x x
8.4 Partial Fractions
3 2
2
2 4 3
2 3
x x x
x x
If the degree of the numerator is higher than the degree of the denominator, use long division first.
2 3 22 3 2 4 3x x x x x 2x
3 22 4 6x x x 5 3x
2
5 32
2 3
xx
x x
5 3
23 1
xx
x x
3 2
23 1
xx x
(from example one)
8.4 Partial Fractions
22
2 4
1 1
x
x x
irreduciblequadratic
factor
repeated root
22 1 1 1
Ax B C D
x x x
2 2 22 4 1 1 1 1x Ax B x C x x D x
2 3 2 22 4 2 1 1x Ax B x x C x x x Dx D
3 2 2 3 2 22 4 2 2x Ax Ax Ax Bx Bx B Cx Cx Cx C Dx D
8.4 Partial Fractions
a
x
2 2a x24
dx
x
These are in the same form.
2
24 x
24sec
2
x
22sec 4 x
tan2
x
2 tan x
22sec d dx
22sec
2sec
d
sec d ln sec tan C
24ln
2 2
x xC
8.4 Partial Fractions
24
dx
x
22sec
2sec
d
sec d ln sec tan C
24ln
2 2
x xC
24ln
2
x xC
2ln 4 ln 2x x C This is a constant.
2ln 4 x x C
8.4 Partial Fractions
a
x
2 2a x
If the integral contains ,
we use the triangle at right.
2 2a x
If we need , we
move a to the hypotenuse.
2 2a x If we need , we
move x to the hypotenuse.
2 2x a
a
x
2 2a x
a
x2 2x a
8.4 Partial Fractions
2
29
x dx
x 3
x
29 x
sin3
x
3sin x
3cos d dx
29cos
3
x
23cos 9 x 29sin 3cos
3cos
d
1 cos 2
9 2
d
9
1 cos 2 2
d
9 9 1sin 2
2 2 2C
sin3
x 1sin3
x
219 9 9
sin2 3 2 3 3
x x xC
8.4 Partial Fractions
2
29
x dx
x
3
x
29 x
19 9sin 2sin cos
2 3 4
xC
219 9 9
sin2 3 2 3 3
x x xC
1 29sin 9
2 3 2
x xx C
9 9 1sin 2
2 2 2C
8.4 Partial Fractions
22
dx
x x We can get into the necessary form
by completing the square.
22x x
22x x
2 2 x x
2 2 1 1x x
21 1x
21 1x
sin u
cos d du 2
1 1
dx
x
8.4 Partial Fractions
22
dx
x x 2
1 1
dx
x
Let 1u x
du dx
21
du
u 1
u
21 ucos
cos
d
d C 1 sin u C
1 sin 1x C
21cos
1
u 21 u
sin u cos d du
8.4 Partial Fractions
24 4 2
dx
x x Complete the square:24 4 2x x
24 4 1 1x x
22 1 1x
22 1 1
dx
x
8.4 Partial Fractions
24 4 2
dx
x x
22 1 1
dx
x
Let 2 1u x
2 du dx
2
1
2 1
du
u
1
u
2 1u
2
2
1 sec
2 sec
d
1
2d
1
2C
11 tan
2u C 11
tan 2 12
x C
tan u
2sec 1u
1
2du dx
2sec d du
2 2sec 1u
8.4 Partial Fractions
Here are a couple of shortcuts that are result from Trigonometric Substitution:
12 2
1tan
du uC
u a a a
1
2 2sin
du uC
aa u
These are on your list of formulas. They are not really new.
8.4 Partial Fractions