来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/07/20 03:32:04
求关于函数极限的题2
![](http://img.wesiedu.com/upload/f/83/f836ff7cbb1765b5abf47fc743d7a520.jpg)
2题 是 X的3次方.
原式=lim(x->1)(3x²-3)/(3x²-5)
=(3-3)/(3-5)
=0
(4)原式=lim(x->+∞)(2lnx·1/x)/1
=lim(x->+∞)2lnx/x
=lim(x->+∞)(2/x)/1
=0
(6)原式=lim(x->0)(1-1/(x+1))/2x
=lim(x->0)(x/(x+1))/2x
=lim(x->0)(1/(x+1))/2
=1/2