x→-1时,
lim(1+1/x)ln(1+x)
=lim[(x+1)ln(1+x)]/x
=-lim[(x+1)ln(1+x)]
=-limln(1+x)/[1/(x+1)]
应用洛必达法则
=-lim[1/(1+x)]/[-1/(x+1)^2]
=lim[1/(1+x)]/[1/(x+1)^2]
=lim x+1
=0
x→-1时,lim(1+1/x)ln(1+x)右极限是0