IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - very good limit question-2

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Numerator, N = x(y²-(y-x)²)^1/2

N = x(y²-(y²+x²-2xy))^1/2

N = x(2xy-x²)^1/2

N = x^3/2(2y-x)^1/2

Denominator, D = {(8xy-4x²)^1/2 + (8xy)^1/2}³

Taking x common : D = x^3/2{(8y-4x)^1/2 + (8y)^1/2}³

Limit, L = _[x]

_[0] x^3/2(2y-x)^1/2 / x^3/2{(8y-4x)^1/2 + (8y)^1/2}³

L = _[x]

_[0] (2y-x)^1/2 / {(8y-4x)^1/2 + (8y)^1/2}³

L = (2y)^1/2 / {(8y)^1/2 + (8y)^1/2}³ = 1/128y

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