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Differential Calculus

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28 Dec 2011 22:47:22 IST

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limits

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plz ans this ques frndz

given limit x tending to infinity xf(x)/((x^5)+1) =1,f(1)=1,f(2)=2,f(3)=3,f(4)=4,then value of f(5)+1/6

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rohit nimmagadda

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Joined: 25 Dec 2011

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29 Dec 2011 22:47:53 IST

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for the limit to exist f(x) should be x^5 +bx^4 +cx^3 +dx^2 +ex +f now i think we have to substitte the valuse given in the guestion and solving all the 4 equations we get the answer

saurav

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30 Dec 2011 02:55:18 IST

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For the limit condition to be true f(x) should be a polynomial with coefficient of x^4=1.next we can see that the roots of f(x)-x are 1,2,3 and 4.so f(x)-x=(x-1)(x-2)(x-3)(x-4).so put x=5 in the above equation to get answer f(5)+1/6=5

Krishna Gopal Singh

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26 Jan 2012 10:49:04 IST

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Limit for xf(x)/(X^5+1) = 1 at infinity so f(x) is a polynomial of degree 4.

From four values of f(x) we get f(x) -x is zero for x = 1,2,3,4

So f(x) = (x-1)(x-2)(x-3)(x-4) +x.

So f(5) = 4*3*2*1+5 = 29

So f(5)+1/6 = 29 +1/6

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