C A L C U L U S

I N D E T E R M I N AT E F R O M S&

I M P R O P E R I N T I G R A LB.E. SEM 1

ELECTRICAL DIPARTMENTDIVISION K

GROUP(41-52)

Zero divided by zero can not be evaluated. The limit may or may not exist, and is called an indeterminate form.

2

2

4lim

2x

x

x

Consider: or

If we try to evaluate by direct substitution, we get:0

0

In the case of the first limit, we can evaluate it by factoring and canceling:

2

2

4lim

2x

x

x

2

2 2lim

2x

x x

x

2lim 2x

x

4

1

lnlim

1 x

x

x

This method does not work in the case of the second limit.

Indeterminate forms

INDETERMINATE FORMS

The expressions of the form

which all are undefined and are called

Indeterminate forms.

1,0,,0,,,0

0 00

L’HÔPITAL’S RULE

Suppose that ( ) ( ) 0, that '( ) and '( ) exist, and that

( ) '( )'( ) 0. Then lim .

( ) '( )x a

f a g a f a g a

f x f ag a

g x g a

Example:

20

1 coslimx

x

x x

0

sinlim

1 2x

x

x

0

If it’s no longer indeterminate, then STOP differentiating!

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!

INDETERMINATE FORM

Evaluate

3

1

6

2

6

22

35

22

2

Lt

Lt

Lt

x

x

x

formx

x

formx

xx

1lim sinx

xx

This approaches0

0

1sin

lim1x

x

x

This approaches 0

We already know that0

sinlim 1x

x

x

but if we want to use L’Hôpital’s rule:

2

2

1 1cos

lim1x

x x

x

1sin

lim1x

x

x

1lim cosx x

cos 0 1

INDETERMINATE FORM

Rewrite as a ratio!

0

1

1 1lim

ln 1x x x

If we find a common denominator and subtract, we get:

1

1 lnlim

1 lnx

x x

x x

Now it is in the form0

0

This is indeterminate form

1

11

lim1

lnx

xx

xx

L’Hôpital’s rule applied once.

0

0Fractions cleared. Still

1

1lim

1 lnx

x

x x x

INDETERMINATE FORM

Rewrite as a ratio!

1

1lim

1 1 lnx x

L’Hôpital again.1

2Answer:

Indeterminate Forms: 1 00 0Evaluating these forms requires a mathematical trick to change the expression into a ratio.

ln lnnu n uln1u

n

We can then write the expression as a ratio, which allows us to use L’Hôpital’s rule.

limx a

f x

ln limx a

f xe

lim lnx a

f xe

We can take the log of the function as long as we exponentiate at the same time.

Then move the limit notation outside of the log.

Indeterminate Powers

INDETERMINATE FORM 00

0

Find lim .xxx

0

0

The limit is of the indeterminate form 0 . Let ( ) and take the

logarithm of both sides. ln ( ) ln

ln ln

1

ˆApply l'Hopital's Rule: lim

x

x

x

f x x

f x x

xx x

x

0

0

2

0

ln ( )

0 0 0

lnln ( ) lim

1

1

lim1

lim( ) 0.

Therefore, lim lim ( ) lim

x

x

x

x f x

x x x

xf x

x

x

xx

x f x e

0 1e

INDETERMINATE FORM ∞0

1

Find lim .xxx

1 lnLet ( ) . Then ln ( ) .

lnˆApply l'Hopital's Rule to ln ( ) : lim ln ( ) lim

1

lim1

x

x x

x

xf x x f x

xx

f x f xx

x

1ln ( ) 0

1 lim 0

Therefore, lim lim ( ) lim 1.

x

f xx

x x x

x

x f x e e

INDETERMINATE FORM 1∞

2

2

ln ( ) 1

ˆApply l'Hopital's Rule:

1ln 1

lim ln ( ) lim1

1 11

1 lim

1

1 lim 1

11

1Therefore, lim 1 lim ( ) lim .

x x

x

x

x

f x

x x x

xf x

x

xx

x

x

f x e e ex

IMPROPAR INTIGRAL

IMPROPAR INTIGRAL

TYPE-1INFINTE

INTERVALS

TYPE-2DISCONTINUOS

INTEGRANDS

DEFINITION OF AN IMPROPER INTEGRAL OF TYPE 1

DEFINITION OF AN IMPROPER INTEGRAL OF TYPE 1

DEFINITION OF AN IMPROPER INTEGRAL OF TYPE 2

DEFINITION OF AN IMPROPER INTEGRAL OF TYPE 2

A COMPARISON TEST FOR IMPROPER INTEGRALS

Sometimes it is impossible to find the exact value of an improper integral and yet it is important to know whether it is convergent or divergent. In such cases the following theorem is useful. Although we state it for Type 1 integrals, a similar theorem is true for Type 2 integrals.