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Xavier4

Find the integer 'n' for which the the

is finite non-zero number.

Conjurer

Apply expansion to all the terms :)

animal

hey i have tried to solve it plz see my soln in the diffrential calculus forum and tell me if i m correct

budokai_tenkaichi_returns

ans>>

lim x to 0 , [cosx(cosx -1) - e^x( cosx-1) ]/ xⁿ -x³/2xⁿ

=(cosx-1)(e^x-1)/xⁿ -x³/2xⁿ

=(2sin²x/2)(e^x-1)/xⁿ -x³/2xⁿ

apply 2 of these >

1. lim x to 0 ,, sinx/x =1

2. lim x to 0 ,, (e^x-1)/x = 1

that simplifies your answer ..

=(1/2)(1)(1)/x^n-3 - 1/2(x^3-n)

=1/2x^3-n- 1/2x^3-n

=0 ..........for n>3

??????????? but you asked for a non zero ..?? something fishy>>

ankitagg

that should be (cos x - e^x).how did u get (e^x-1)

budokai_tenkaichi_returns

yes ..u r right ....thanx...here we have to apply expansion of cosx and ex ..that sure wil do it ..

ankurgupta91

= [cosx(cosx-1) + e^x ( 1-cosx)]/x^n - x^3/2x^n

= [(1-cosx)(e^x-cosx)]/x^n - 1/2x^(n-3)

= [2sin^2(x/2)(e^x-cosx)] / (x^2)(x^n-2) -1/2x^(n-3)

as lim x------>0 sinx/x = 1
so
= (1/2) [ e^x - cosx - x ] / x^(n-2)

its nw 0/0 form we can apply l-hospital rule
so,
= (1/2) [ e^x +sinx - 1] / [(n-2)[x^(n-3)]]

= (1/2(n-2)) [ [[e^x-1]/x] + sinx/x ] / x^(n-4)
by apllying limits

= (1/2(n-2)) [ 2/x^(n-4)]

= 1/ [ (n-2) ( x^(n-4)]
so for existence of limit
n=4
nd then we get the ans = 1/2

thats the answer
hope u all gt it.........

amitp91

amitp91

whats the rite ans

supriyaraj

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