Intermediate Forms and L’Hospital Rule (also L’Hopital and Bernoulli Rule)

Historically, this result first appeared in L'Hôpital's 1696 treatise, which was the first textbook on differential calculus.

Within the book, L'Hôpital thanks the Bernoulli brothers for their assistance and their discoveries.

An earlier letter by John Bernoulli gives both the rule and its proof, so it seems likely that Bernoulli discovered the rule.

Definition: Indeterminate Limit/Form

If exists, where , the limit is said to be indeterminate.

lim𝑥→𝑐

𝑓 (𝑥)𝑔(𝑥 )

The following expressions are indeterminate forms:

These expressions are called indeterminate because you cannot determine their exact value in the indeterminate form.

However, it is still possible to solve these in many cases due to L'Hôpital's rule.

We used a geometric argument to show that:

You have previously studied limits with the indeterminate form

I Indeterminate Form

Example:

Example: 42222

)2)(2(24

limlimlim22

2

2

x

xxx

xx

xxx

Example:

xxx

xxx

xx

xxx 2sin1

3cos1

13sin

2sin3cos3sin

2sin3tan limlimlim

000

Example:

23)1)(1)(1(

23

2sin2

3cos1

33sin

23 limlimlim

02003

xx

xxx

xxx

Some limits can be recognized as a derivative

hafhafaf

h

)()()( lim0

Recognizing a given limit as a derivative (!!!!!!)

Example:

Example:

Example:

Tricky, isn’t it? A lot of grey cells needed.

Not all forms are like those.

with the knowledge given to you

by now you get a doctorate at 17 and get to quit the school right now.

If you can find

L’Hospital rule for indeterminate form

Let f  and g  be real functions which are continuous on the closed interval [a, b]  and differentiable on the open interval (a, b)  . Suppose that . Then:

lim𝑥→𝑐

𝑓 (𝑥)𝑔(𝑥 )

=lim𝑥→𝑐

𝑓 ′ (𝑥)𝑔 ′ (𝑥)

provided that the second limit exists.

Example:

Example: 4)2(21

224

limlim2

2

2

xxx

xx

23

)1(2)1(3

2cos23sec3

2sin3tan 2

00limlim

xx

xx

xxExample:

121

)8(3

1

)8(3

11

)1()8(31

283

23

20

32

0

3

0limlimlim

h

h

hh

hhhExample:

21

2211

limlimlim00

20

x

x

x

x

x

x

ex

exxe

Example:

23

3sin1

sin

3

21cos

limlim33

xx

x

xxExample:

010

1cos

2

11cos

2

1sin

1limlimlim

2

32

x

x

xx

x

x

xxxx

Example:

Example: 0

1)0(2

cos2

sin1sin

1limlimlim

0

2

0

2

yy

yy

x

xyyx xy 1

Example:

Example: lim𝑥→ 2

sin (𝑥2−4 )𝑥−2

=lim𝑥→2

2𝑥 cos (𝑥2−4 )1

=4

Suppose that instead of , we have that   and   as . Then:

Corollary for indeterminate form

lim𝑥→𝑏−

𝑓 (𝑥)𝑔 (𝑥)

= lim𝑥→𝑏−

𝑓 ′ (𝑥)𝑔 ′ (𝑥)

provided that the second limit exists

II Indeterminate Form

23

46

3456

132753 limlimlim 2

2

xxx xx

xxxx

Example:

Easier: 23

002003

132

753

132

753

132753 limlimlimlim

2

2

222

2

222

2

2

2

xxxx

xx

xx

xxx

xx

xxx

xx

xxxx

=

418

49

1243 limlimlim

2

2

3 xxx

xx

xxxExample:

1243

2

3

lim xx

x

03

0003

12

43

3

3

limxx

xx

limit does not exist.

3

23322

1312

13

12

)1ln()1ln(

24

4

22

3

3

2

2

3

2

limlimlimlim

xxxx

xxxx

xx

xx

xx

xxxxExample:

02

0222

1 1

ln 22

0

3

03020limlimlimlim

xx

x

x

x

x

xxxxx

=

=

=Example:

Example:

14

41

142

8

114

2

22

limlimlimx

xx

x

xx

xxx

Rule does not help in this situation. It is a pure pain. In the situations like this one divide in you mind denominator and numerator with highest exponent of

Example:

2/11/14

114 limlim

2

x

xxx

xx

III Indeterminate Form

0 ∙ ∞ use algebra to convert the expression to a fraction (0 ), and then apply L'Hopital's Rule

Example: lim𝑥→𝜋 /2−

(𝑥− 𝜋2 ) tan𝑥= (0 ∙∞ )= lim

𝑥→ 𝜋/2−

𝑥− 𝜋2cot𝑥 =( 0

0 )= lim𝑥→𝜋 /2−

1−𝑐𝑠𝑐2𝑥

=−1

Example: 0)(1

1

1lnln limlimlimlimlim

0

2

02000

x

xx

x

x

x

xxxxxxxx

x

xxxx

xxxxx

xxxx

tansincotcsc

1

csclnln)(sin limlimlimlim

0000

Example:

0)0)(1(tansin limlim00

xxx

xx

Example: 1sin

1

1sin1sin limlimlim

0

yy

x

xxx

yxx xy 1

Example:

Example:

A limit problem that leads to one of the expressions

is called an indeterminate form of type (+∞ )− (+∞ ) ,   (−∞ )− (−∞ ) ,   (+∞ )+(−∞ ) ,   (−∞ )+(+∞ )

Such limits are indeterminate because the two terms exert conflicting influences on the expression; one pushes it in the positive direction and the other pushes it in the negative direction

IV Indeterminate Form

Indeterminate forms of the type can sometimes be evaluated by combining the terms and manipulating the result to produce an indeterminate form of type

00 𝑜𝑟

∞∞

Example: convert the expression into a fraction by rationalizing

xxxx

xxxx

xx xxx sincos1cos

sinsin

sin11

limlimlim000

Example:

020

coscossinsinlim

0

xxxxx

x

Example:

20

2

0

cos1lnln)cos1ln( limlim xxxx

xx

2

1ln2

sinlncos1ln limlim0

20 x

xxx

xx

Example:

Example:

V Indeterminate Form

Several indeterminate forms arise from the limit

These indeterminate forms can sometimes be evaluated as follows:

)()( xgxfy

)(ln)()(lnln )( xfxgxfy xg

)(ln)(ln limlim xfxgyaxax

1.

2.

3.

Take of both sides

Find the limit of both sides

)(ln)(lim xfxgax

4. If = L

5.

Example: 1∞

ln y = ln[(1 + sin 4x)cot x] = cot x ln(1 + sin 4x)

FindExample: 00

l n 𝑦=𝑥 ln 𝑥lim

𝑥→ 0+¿ ln𝑦= lim𝑥→0 +¿𝑥 ln𝑥=(0 ∙−∞ )= lim

𝑥→ 0+ ¿ ln𝑥1𝑥

= lim

𝑥→ 0+ ¿

1𝑥

− 1

𝑥2

= lim𝑥→ 0+¿ (− 𝑥 )=0

¿

¿ ¿

¿ ¿¿ ¿¿ ¿ ¿¿

¿

lim𝑥→0+¿𝑦= lim

𝑥→0+ ¿𝑒 ln𝑦=𝑒0=1¿ ¿¿

¿

xx = (eln x)x = ex ln x

lim𝑥→0+¿𝑥𝑥¿

¿

lim𝑥→ 0+¿𝑥𝑥=1¿

¿

xx

xe

2)1(lim

Example: Find 0

xxey

2)1(

xeeyx

xx )1ln(2)1(lnln2

xeyx

xx

)1ln(2ln limlim

22 1

2 1

12

limlimlim

x

x

xx

x

x

x

x

x ee

eee

e

xx

xe

2)1(lim

2e

(∞∞ )

(∞∞ )

FindExample: x

xx

1

0coslim

1

xxy1

)(cos

xxxy x )ln(cos)(coslnln

1

x

xyxx

)ln(cosln limlim00

( 00 )

0tanlim0

xx

x

xx

1

0coslim

10 e