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Differential Calculus

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14 Jan 2007 21:35:59 IST

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780

LIMITS

None

Sir,

Plz explain how to solve.

If limit x tending to zero of

(acosx+bxsinx-5)/x⁵ is a finite value then (a,b)=?

Ans is given as (5,5/2)

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Comments (5)

Manasi

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14 Jan 2007 21:50:17 IST

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since the value of this limit is infinite, then it must be of 0/0 form, in order to apply L'Hospital Rule... that means Nr=0

acosx+bxsinx-5 = 0 ..............(1)

Now differentiating the expression once and equating it with 0 again,we have,

-asinx+bxcosx+bsinx=0

differentiating once again,

-acosx-bxsinx+2bcosx=0 ................(2)

Solving 1 and 2,

2bcosx=5,

which is possible only when cosx=1,

hence b=5/2 and a=5 (be using the values in eqn (1))

Hopefully u'll get it now..... any problem do ask again.......

CyBorG

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15 Jan 2007 16:40:22 IST

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You are right.But 2bcosx=5 therefore b=5/2 since x tends to zero and cosx tends to 1 which is not the reason given by you.

Manasi

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15 Jan 2007 21:37:49 IST

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yupp u r rite, cos x= 1as x tends to 0..... thanx ....

newayz, u got the solution, thats gud!!!

deepak_agarwal

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17 Jan 2007 09:07:00 IST

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the ques is a simple application of L'Hospitals rule....since the limit exists and denominator approach 0 as x tends to 0 so nt must also approach zero....if u diff num and den then also the same case tht den approach zero......thats what is done by magiclko

vijay kharya

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Joined: 27 Aug 2009

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14 Jun 2010 09:21:35 IST

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an even better approach would be the application of series expansions!

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