lim[1/(x^2+x)]
=lim1/[x(x+1)]
=lim[1/x-1/(x+1)]
=lim1/x-lim1/(x+1)
当x→0时,lim1/x=∞,lim1/(x+1)=1
所以lim[1/(x^2+x)]=∞ （当x→0时）