4.5 - Indeterminate Forms and L’Hospital’s Rule

“For

€

limx→a

f (x)

g(x)where both

€

f (x) → 0 and g(x) → 0 as x → a

then the limit may or may not exist and is called an indeterminate form of type ”

€

0

0

L’Hospital’s Rule: “For

€

limx→a

f (x)

g(x)where IDF

€

0

0 or

€

±∞±∞ result

€

limx→a

f (x)

g(x)= limx→a

f '(x)

g'(x)

IF the limit on the right side exists or is

then

€

=±∞

ex:

€

limx→1

ln x

x −1=

0

0 →H

limx→1

d

dxln x

d

dxx −1

€

=limx→1

1/ x

1 = 1

€

→H

limx→∞

d

dxex

d

dxx 2

= ex

2x =

∞

∞

€

→H

limx→∞

d

dxex

d

dx2x

= ex

2 = ∞

ex:

€

limx→∞

ex

x 2 =

∞

∞

ex:

€

limx→∞

ln x

x3 =

∞

∞

€

→H

limx→∞

1/ x

1

3x−2 / 3

= 0

0

€

1

x1

3x 2 / 3

= 1

x •

3x 2 / 3

1 =

3x 2 / 3

x = 3x−1/ 3 =

3

x3

Simplify

€

limx→∞

3

x3 = 0

ex:

€

limx→0

tan x − x

x 3 =

0

0

€

→H

limx→0

sec2 x −1

3x 2 =

0

0

€

→H

limx→0

2sec2 x tan x

6x =

0

0

€

→H

limx→0

4 sec2 x tan2 x + 2sec4 x

6 =

2

6 =

1

3

ex:

€

limx→π −

sin x

1− cos x

€

= 0

1− (−1) = 0

HW 4.5A - pg 308#’s 1-10 all

HW 4.5A - pg 308#’s 1-10 all

Indeterminate Products

Where

€

limx→af (x) = 0 and lim

x→ag(x) = ∞

It is not clear what the value of

€

limx→af (x)g(x) if any, will be.

This limit is called an indeterminate form of type

€

0 • ∞

Solution: convert it to a quotient….

ex:

€

fg =f

1/g or fg =

g

1/ f

ex: xxx

lnlim0+→

Indeterminate Differences

Where ∞=∞=→→

)(lim )(lim xgandxfaxax

Then [ ])()(lim xgxfax

−→

is indeterminate form of type ∞−∞

Solution: find a common denominator….

ex: xxx

tanseclim2/

−→π

writeRe

→∞−∞=x

x

xx cos

sin

cos

1lim

2/−

→π

x

xx cos

sin1lim

2/

−=

→π 0

0=

x

xx

H

sin

coslim

2/ −−

→→π

0=

Indeterminate Powers

3 types:

€

limx→a

f (x)[ ]g(x )

For

00)1

0)2 ∞∞1)3

• Solve by either ln’ing both sides OR exponentiating.

•Will give us ∞•0 type IDF.

•ex: [ ] x

x

x cot

0

4sin1lim ++→

∞= 1

[ ][ ]

[ ]IDFy

xxy

xy

xyletx

x

0ln

4sin1lncotln

4sin1lnln

4sin1cot

cot

•∞=

+=

+=

+=

[ ]0

0

tan

4sin1lnlim

0

=+

+→ x

x

x

( )[ ] ( )4

114

sec4sin14cos4

limsec

44cos4sin1

1

lim2

02

0

==+=••

+→

→+→ xxx

x

xx

xx

H

4lnlim0lnlim00

=→•∞=+→+→

yIDFyxx

4ln ee y = yx +→

=0

lim

•ex: x

xx

+→ 0lim 00= Solution: Any variable x can be written as elnx

… so we write

€

limx→0+

x x = limx→0+

eln x( )

x

… multiply the exponents xx

xe ln

0lim

+→=

IDFtypexasxx )0(00ln... −∞•→→ 10 ==e

HW 4.5B – pg 308#’s 15, 16, 20, 21-23

all, 25-27 all

!! BONUS !!FOR +5

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