If limx→0 ax−e4x−1axe4x−1=b exists finitely then 2(a+b)=

limx→0 ax−e4x−1axe4x−1=b exists finitely⇒ limx→0 a−4e4x−14xae4x−1=b  exists finitelyThe numerator becomes a-4 when x→0 and the denominator tends to zero. Therefore, if a -4≠0, LHS→∞ whereas RHS is finite. Therefore, a -4 = 0 i.e. a = 4.When a=4 limx→0 a−4e4x4x−1ae4x−1=b⇒ limx→0 1−e4x−14xe4x−1=b ⇒ limx→0 1−1+(4x)2!+(4x)23!+…4x+(4x)22!+(4x)33!+…=b⇒−12=b∴ 2(a+b)=24−12=7