![](https://iknow-base.cdn.bcebos.com/lxb/notice.png)
lim(x→1)sin^2πx/(x-1)lnx 怎么算,求高数大神详细指导
3个回答
展开全部
算法之一如下:
lim(x->1)(sin²πx)/[(x-1)lnx]
=lim(x->1)[sin²(π-πx)]/[(x-1)ln(1+x-1)]
=lim(x->1)(π-πx)²/[(x-1)(x-1)]
=lim(x->1)π²(1-x)²/(x-1)²
=π².
lim(x->1)(sin²πx)/[(x-1)lnx]
=lim(x->1)[sin²(π-πx)]/[(x-1)ln(1+x-1)]
=lim(x->1)(π-πx)²/[(x-1)(x-1)]
=lim(x->1)π²(1-x)²/(x-1)²
=π².
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询