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Differential Calculus

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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

lim

o In(cosx)/x²

^{without using l,hosptial rule}

Comments (7)

vasanth_mech pilani

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18 Feb 2007 19:05:37 IST

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is the answer -1/2

vasanth_mech pilani

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18 Feb 2007 19:10:23 IST

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when x

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

Umang

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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 !

puneet

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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)/x²= 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 x

o [log ( 1 + (-2sin²(x/2)) )/(-2sin²(x/2))] = 1

and, lim x

o [(-2sin²(x/2))/x²] = -1/2

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

cheers

Manan Mahajan

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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......

Rajat Barve

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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

rishikesh singh

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19 Mar 2007 01:57:13 IST

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ln(cosx)/x^{2
(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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