If Ltx→0x1+acosx−bsinxx3=1 then b=

Given Ltx→0x1+acosx−bsinxx3=1Use LH rule to get the finite limitLtx→01+acosx+x−asinx−bcosx3x2=1⇒1+a−b=0⇒a−b=−1Again LH ruleLtx→0−asinx+−asinx+x−acosx+bsinx6x=1 Ltx→0−2a+bsinx+x−acosx6x=1Ltx→0−2a+b6⋅sinxx−acosx6=1−2a+b6−a6=1−3a+b=6Multiply the equation a−b=−1 with 3 and then add to the equation −3a+b=6it gives 3a−3b−3a+b=−3+6⇒−2b=3⇒b=−32