各位大神 帮忙给做一下这个题吧。谢谢了。

Question

如题所述

Answer 1

要知道一点
limsinx=x（其中x趋向于0）
limsin(x+3)=x+3
原式=lim[(x2+2x-3)/(x+3)]=lim[(x+3)2-4(x+3)]/(x+3)
=limx+3-4=limx-1=-4
满意请采纳

Answer 2

学过洛必塔法则就用法①

否则利用重要极限计算

Answer 3

lim(x->-3) (x^2+2x-3)/sin(x+3)
let
y=x+3
lim(x->-3) (x^2+2x-3)/sin(x-3)
=lim(y->0) ((y-3)^2+2(y-3)-3)/siny
=lim(y->0) ((y-3)^2+2(y-3)-3)/y
=lim(y->0) (y^2-4y)/y
=lim(y->0) (y-4)
=-4