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Differential Calculus

Forum Expert
 Joined: 18 Feb 2007 Post: 67
18 Feb 2007 18:23:52 IST
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limits
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Differential Calculus

calculate the limit of the folllowing expression

limo In(cosx)/x

without using l,hosptial rule

Blazing goIITian

Joined: 4 Feb 2007
Posts: 526
18 Feb 2007 19:05:37 IST
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Blazing goIITian

Joined: 4 Feb 2007
Posts: 526
18 Feb 2007 19:10:23 IST
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when x0
the expression ln(cosx)/x2 = 0/0 (indeterminate form)

therefore by l'hospital's rule
=lim [(1/cosx).(-sinx)]/2x
=lim [-1/2cosx] . lim (sinx/x)
=-1/2(1)
= -1/2

if u kno any other method without using l'hospitals rule plz share it here

Blazing goIITian

Joined: 12 Dec 2006
Posts: 926
18 Feb 2007 19:57:55 IST
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Hey vasanth !
do u know any method other than this to solve log limits ???
If yes , then pls share !

Forum Expert
Joined: 19 Oct 2006
Posts: 1966
18 Feb 2007 20:46:40 IST
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hii

this one is simple .. lets see how to do this ..

well .. log(cosx)/x= log( 1 + (cosx - 1) )/x2
= log( 1 + (-2sin2(x/2)) )/x2
= [log ( 1 + (-2sin2(x/2)) )/(-2sin2(x/2))].[(-2sin2(x/2))/x2]

Now we know lim x->0 log( 1 + x ) / x = 1

So, lim xo [log ( 1 + (-2sin2(x/2)) )/(-2sin2(x/2))] = 1

and, lim xo [(-2sin2(x/2))/x2] = -1/2

so, the answer is 1.(-1/2) = -1/2

cheers

Cool goIITian

Joined: 16 Mar 2007
Posts: 39
16 Mar 2007 08:37:18 IST
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the most simple method is:
logcosx=log(1+cosx-1)which can be expanded using expansion of ln(1+x)=x-x^2/2---------------
x^/2 are neglected as they would always tend to zero.
hence ln(1+cosx-1)can be simply written as=(cosx-1)
therefore limit simplifies to:
Lt (cosx-1)/x^2which is -1/2..........
Hope you understand......

Blazing goIITian

Joined: 31 Jan 2007
Posts: 780
16 Mar 2007 09:21:32 IST
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i think the most simple method is this,
multiply and divide by 2 ,the given expression

therefore z = 2log(cosx)/2x^2
=> z = log(cos^2x)/2x^2
cos^x can be written as 1 - sin^2x
therefore log(cos^2x) = log (1 - sin^2x)
this can be expanded by using log (1-x)
so the expansion is -(sin^2x) + (sin^4x)/2 + ................
when this is divided by x^2 and the limit ttaken,the first term equals 1 and others  = 0
therefore z = -1/2  ...................(remember we multiplied and divided by 2 )

all those who are satisfied plzzz. rate me

Cool goIITian

Joined: 19 Mar 2007
Posts: 96
19 Mar 2007 01:57:13 IST
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ln(cosx)/x2
(i have used some rule which are simple to prove and can be used to simplify calculation)
= ln(1-2sin2x/2)/x2
rules:  [ x][0 ] sinx tends to x
=ln(1-x2/2)/x2
rule :[t ][ o] ln(1+t) tends to t
here x tend to 0 so x2 tends to 0 so simply replace ln(1-x2/2) by -x2/2

hence (-x2/2)/x2
= -1/2

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