Question

求下列极限题目过程谢谢

Answer 1

(1)分母→0，极限存在，分子→0,0/0，用洛必达法则 ∴原式=lim(x→0)(1+sin2x)/1=1 (2)=lim(x→0)arcsinx/2x=lim(x→0)1/2√(1-x2)=? (3)=lim(x→0)(1-e^x2)/(2xsin2x+2x2cos2x) =lim(x→0)(-2xe^x2)/(2sin2x+4xcosx+4xcos2x-4x2sin2x) =lim(x→0)(-e^x2)/(2sin2x/2x+4cosx-2xsin2x) ∵lim(x→0)sin2x/2x=1 ∴原极限=-1/6 (4)为∞/∞用洛必达法则 =lim(x→+∞)[2∫(0,x)e^u2du)·e^x2]/e^2x2 =lim(x→+∞)[2∫(0,x)e^u2du)]/e^x2 =lim(x→+∞)[2e^x2)]/2xe^x2 =lim(x→+∞)2/2x =0

追问

能否不用到洛必达，没学到