The value of a so that limx→0 1x2eαx−ex−x=32 is

We have,limx→0 eαx−ex−xx2=32⇒limx→0 1+αx+(αx)22!+…−1+x+x22!+…−xx2=32⇒     limx→0 (α−2)x+α2−1x22!+x33!α3−1+…x2=32⇒     α−2=0⇒α=2