If limx→ax5−a5x−a=405, find all possible values of a.
Given limx→ax5−a5x−a=405....(i)
LHS=limx→ax5−a5x−a
=5(a)5−1
=5a4 It is given that 5a4=405
⇒5a4=405
a4=4055=81
a4=(3)4,a2=9
a=±3
⇒a=3 and a=-3
If limx→ax9−a9x−a=limx→5(4+x), find all possible values of a.
If limx→ax9−a9x−a=9, find all possible values of a.