lim x→0 ex?(ax2+bx+1) x2 = lim x→0 ex?2ax?b 2x ∵ lim x→0 2x=0∴ lim x→0 (ex?2ax?b)=1?b=0∴b=1∴原式= lim x→0 ex?2ax?b 2x = lim x→0 ex?2a 2 = 1?2a 2 =0∴a= 1 2 故选:A.