B.E. SEM 1 ELECTRICAL DIPARTMENT DIVISION K GROUP(41-52) B.E. SEM 1 ELECTRICAL DIPARTMENT DIVISION K...
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Transcript of B.E. SEM 1 ELECTRICAL DIPARTMENT DIVISION K GROUP(41-52) B.E. SEM 1 ELECTRICAL DIPARTMENT DIVISION K...
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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)
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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
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INDETERMINATE FORMS
The expressions of the form
which all are undefined and are called
Indeterminate forms.
1,0,,0,,,0
0 00
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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
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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!
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INDETERMINATE FORM
Evaluate
3
1
6
2
6
22
35
22
2
Lt
Lt
Lt
x
x
x
formx
x
formx
xx
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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
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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:
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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
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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
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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
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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
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IMPROPAR INTIGRAL
IMPROPAR INTIGRAL
TYPE-1INFINTE
INTERVALS
TYPE-2DISCONTINUOS
INTEGRANDS
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DEFINITION OF AN IMPROPER INTEGRAL OF TYPE 1
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DEFINITION OF AN IMPROPER INTEGRAL OF TYPE 1
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DEFINITION OF AN IMPROPER INTEGRAL OF TYPE 2
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DEFINITION OF AN IMPROPER INTEGRAL OF TYPE 2
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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.