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(

2

4limlimlim

22

2

2

x

x

xx

x

x

xxx

Example:

xx

x

xxx

x

x

xxx 2sin

1

3cos

1

1

3sin

2sin3cos3sin

2sin

3tanlimlimlim

000

Example:

2

3)1)(1)(1(

2

3

2sin

2

3cos

1

3

3sin

2

3limlimlim

02003

x

x

xx

x

xxx

Some limits can be recognized as a derivative

h

afhafaf

h

)()()( lim

0

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

2

2

4limlim

2

2

2

x

x

x

xx

2

3

)1(2

)1(3

2cos2

3sec3

2sin

3tan 2

00limlim

x

x

x

x

xxExample:

12

1

)8(3

1

)8(3

1

1

)1()8(3

128

32

32

0

32

0

3

0limlimlim

h

h

h

h

hhhExample:

2

1

22

11limlimlim

002

0

x

x

x

x

x

x

e

x

e

x

xeExample:

2

33sin

1

sin

3

21cos

limlim33

x

x

x

xxExample:

01

01cos

2

11cos

2

1sin

1

limlimlim2

32

x

x

xx

x

x

xxxx

Example:

Example:

01

)0(2

cos

2

sin1sin

1

limlimlim0

2

0

2

y

y

y

y

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

2

3

4

6

34

56

132

753limlimlim 2

2

xxx x

x

xx

xxExample:

Easier: 2

3

002

00313

2

753

132

753

132

753limlimlimlim

2

2

222

2

222

2

2

2

xxxx

xx

xx

xxx

xx

xxx

xx

xx

xx

=

4

18

4

9

12

43limlimlim

2

2

3 x

x

x

x

x

xxx

Example:

12

432

3

lim x

x

x

0

3

00

0312

43

3

3

limxx

xx

limit does not exist.

3

2

33

22

13

12

1

31

2

)1ln(

)1ln(24

4

22

3

3

2

2

3

2

limlimlimlim

xx

xx

xx

xx

x

xxx

x

x

xxxxExample:

02

0

222

1

1ln 22

0

3

03020limlimlimlim

x

x

x

x

x

x

x

xxxx

=

=

=Example:

Example:

14

4

1142

8

1

142

22

limlimlimx

xx

x

x

x

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

1

14limlim

2

x

x

x

x

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 ∙∞ )= lim𝑥→ 𝜋/2−

𝑥−𝜋2

cot𝑥=( 0

0 )= lim𝑥→𝜋 /2−

1

−𝑐𝑠𝑐2𝑥=−1

Example: 0)(1

1

1ln

ln limlimlimlimlim0

2

02000

x

x

x

x

x

x

xxx

xxxxx

x

xx

xxx

x

xxx

xxxx

tansin

cotcsc

1

csc

lnln)(sin limlimlimlim

0000

Example:

0)0)(1(tansin

limlim00

xx

x

xx

Example: 1

sin1

1sin1sin limlimlim

0

y

y

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

00𝑜𝑟∞∞

Example: convert the expression into a fraction by rationalizing

xxx

x

xx

xx

xx xxx sincos

1cos

sin

sin

sin

11limlimlim

000Example:

02

0

coscossin

sinlim

0

xxxx

x

x

Example:

20

2

0

cos1lnln)cos1ln( limlim

x

xxx

xx

2

1ln

2

sinln

cos1ln limlim

02

0 x

x

x

x

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+ ¿ ln𝑥

1𝑥

= lim

𝑥→ 0+ ¿

1

𝑥

−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(

x

eey

xxx )1ln(2

)1(lnln2

x

ey

x

xx

)1ln(2ln limlim

22

1

2

1

12

limlimlim

x

x

xx

x

x

x

x

x e

e

e

eee

xx

x

e2

)1(lim2e

(∞∞ )

(∞∞ )

FindExample: x

x

x1

0

coslim

1

xxy1

)(cos

x

xxy x )ln(cos)(coslnln

1

x

xy

xx

)ln(cosln limlim

00( 0

0 ) 0tanlim

0

xx

x

x

x1

0

coslim

10 e