✔ 最佳答案
lim(x->－1)_(x－2)/(x^2+4x－3)
=(－1－2)/(1－4－3)
=1/2
lim(x->2)_(2x+4)/(x－7)
=(4+4)/(2－7)
=－8/5
lim(x->2)_(x^3+3x^2－10x)/(x－2)
=lim(x->2)_(x^2+5)
=9
lim(x->0)_[2－√(x+4)]/x
=lim(x->0)_[2－√(x+4)]*[2+√(x+4)]/｛x[2+√(x+4)]｝
=lim(x->0)_－x/｛x[2+√(x+4)]｝
=lim(x->0)_－1/[2+√(x+4)]
=－1/(2+2)
=－1/4
lim(x->1)_[1/(x－1)－3/(x^3－1)]
=lim(x->1)_(x^2+x+1－3)/(x^3－1)
=lim(x->1)_(x^2+x－2)/(x^3－1)
=lim(x->1)_(x+2)/(x^2+x+1)
=3/4
lim(x->∞)_(2x+1)/(x^2－x+1)
=lim(x->∞)_(2x/x^2+1/x^2)/(1－x/x^2+1/x^2)
=0
lim(x->∞)_√(x^2－2x+10)
=lim(x->∞)_√(x^2－2x+1)
=lim(x->∞)_(x－1)
=∞
lim(x->1)_[4－√(x+15)]/(x^2－1)
=lim(x->1)_[4－√(x+15)]*[4+√(x+15)]/｛(x^2－1)*[4+√(x+15)]｝
=lim(x->1)_(16－x－15)/｛(x^2－1)*[4+√(x+15)]｝
=－lim(x->1)_(x－1)/｛(x^2－1)*[4+√(x+15)]｝
=－lim(x->1)_1/｛(x+1)*[4+√(x+15)]｝
=－1/[2*(4+4)]
=－1/16
2013-11-02 17:00:41 補充：
lim(x->2)_(x^3+3x^2－10x)/(x－2)
=lim(x->2)_(x^2+5x)
=14
lim(x->1)_[1/(x－1)－3/(x^3－1)]
=lim(x->1)_(x^2+x+1－3)/(x^3－1)
=lim(x->1)_(x^2+x－2)/(x^3－1)
=lim(x->1)_(x+2)/(x^2+x+1)
=3/3
=1
