Suppose limx→0ea1x−1a2x2+a3x=1,\lim_{x\to 0}\frac{e^{a_1 x}-1}{a_2 x^{2}+a_3 x}=1,limx→0a2x2+a3xea1x−1=1, where a1,a2,a3a_1,a_2,a_3a1,a2,a3 are given real numbers. Then it is necessarily true that
a1=a2=a3=1a_1=a_2=a_3=1a1=a2=a3=1
a1=a3≠0a_1=a_3\ne 0a1=a3=0
a2=0a_2=0a2=0
a2+a3≠0a_2+a_3\ne 0a2+a3=0
