Question

Tough one!

Answer 1

Imx->0 (asin2x + b log(cosx))/x4 = 1/2 [0/0 form] ,applying L'Hospital rule ,we get

= > limx->0 (2a*sinx*cosx - (b /cosx)*sinx)/ 4x3 = 1/2 => limx->0 (a*sin2x - b*tanx)/ 4x3 = 1/2 [0/0 form],

applying L'Hospital rule again ,we get,

 = > limx->0 (2a*cos2x - b*sec2x) / 12x2 = 1/2

For above limit to exist,Numerator must be zero so that we get [0/0 form] & we can further proceed.

Hence 2a - b =0 => 2a = b ------(A)

limx->0 (b*cos2x - b*sec2x) / 12x2 = 1/2 [0/0 form], applying L'Hospital rule again ,we get,

= > limx->0 b*(-2sin2x - 2secx*secx.tanx) / 24x = 1/2 => limx->0 2b*[-sin2x - (1+tan2x)tanx] / 24x = 1/2

[0/0 form], applying L'Hospital rule again ,we get,

limx->0 2b*[-2cos2x - (sec2x+3tan2x*sec2x)] / 24 = 1/2 = > 2b[-2 -1] / 24 = 1/2 => -6b/24 = 1/2 => b = -2

from (A), we have , 2a = b => 2a = -2 => a = -1

Hence a =-1 & b = -2