显然1/(x-2)=(x^2+2x+4)/(x^3-8)
所以
原极限
=lim(x→2) (x^2+2x-8)/(x^3-8)
=lim(x→2) (x+4)(x-2)/(x^3-8)
=lim(x→2) (x+4)/(x^2+2x+4) 代入x=2
=6/12
=1/2