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Limits (Question)

djchalil said – Thu, 13 Aug 2009 15:15:38 -0000 ( Link )

limit as ‘x’ tends to ‘a negative’ {(modulus of x/a)^3 – [x/a]} where a<0 & [] is greatest int. fn.

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  1. Kunal Wadhwa said – Thu, 20 Aug 2009 13:07:20 -0000 ( Link )

    ans is zero

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  2. sid_du said – Fri, 25 Sep 2009 16:27:31 -0000 ( Link )

    I think ans. is one (1) Soln:- let y = x/a hence, as x ->a () ,y-> 1 () (mod y)3 – (step y) y -> 1() indicates y is a number just less than 1 (0.99999999999999…) hence , (step y ) = 0 (1)3 – 0 = 1 note : ans would have been 0 had it been x—> a (+). Reply if the ans is correct.

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