2024-01-09 23:13
x→0,y=1/x→∞
lim[ln(1+x)/x]=lim[yln(1+1/y)]=limln(1+1/y)^y=lnlim(1+1/y)^y=lne=1
x→0,y=e^x-1→0,x=ln(1+y)
lim[(e^x-1)/sinx]=lim[(e^x-1)/x]=lim[y/ln(1+y)]=1/lim[ln(1+y)/y]=1/1=1