Guillaume De l'Hôpital 1661 - 1704 Actually, L’Hôpital’s Rule was developed by his teacher Johann Bernoulli. De L’Hôpital paid Bernoulli for private lessons,

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


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


2

2

4lim

2x

x

x

limx a

f x

g xAs 2x

f x

g xbecomes:

df

dg

df

dg

dx

dxdf

xdgd


2

2

4lim

2x

x

x

limx a

f x

g x

2

2

4lim

2x

dx

dxd

xdx

2

2lim

1x

x

4


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


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


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!


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


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.


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


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


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.


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


WARNINGWARNINGWARNING

Please temporarily suspend your common sense.


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.


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


“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.


Show that and grow

at the same rate as .

2 5f x x 2

2 1g x x

x


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.


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


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


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.


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


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.


1

0

dx

x1

0lim ln

bbx

0lim ln1 lnb

b

0

1lim lnb b

This integral diverges.

1

0

1lim

bbdx

x


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


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


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.)


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


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


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


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


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


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,


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


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


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


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


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.


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


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)


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


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


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


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


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


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


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


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


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


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


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.


Documents

Guillaume De l'Hôpital 1661 - 1704 Actually, L’Hôpital’s Rule was developed by his teacher Johann Bernoulli. De L’Hôpital paid Bernoulli for private lessons,