![](https://iknow-base.cdn.bcebos.com/lxb/notice.png)
高数求极限,数学高手帮帮忙,要详细的步骤。。谢谢 10
3个回答
展开全部
解法一:(罗必达法)
(1)原式=e^{lim(x->0)[ln(1-x)/x]}
=e^{lim(x->0)[-1/(1-x)]} (0/0型极限,应用罗比达法则)
=e^(-1)
=1/e;
(2)原式=e^{lim(x->0)[ln(1+2x)/x]}
=e^{lim(x->0)[2/(1+2x)]} (0/0型极限,应用罗比达法则)
=e^2
=e²;
(3)原式=e^{lim(n->∞)[2nln((n+2)/(n+1))]}
=e^{lim(n->∞)[2ln((1+2/n)/(1+1/n))/(1/n)]}
=e^{lim(x->0)[2ln((1+2x)/(1+x))/x]} (令x=1/n)
=e^{lim(x->0)[2/((1+2x)(1+x))]} (0/0型极限,应用罗比达法则)
=e^2
=e²。
解法二:(重要极限法)
(1)原式=lim(x->0){[(1+(-x))^(1/(-x))]^(-1)}
={lim(x->0)[(1+(-x))^(1/(-x))]}^(-1)
=e^(-1) (应用重要极限lim(z->0)[(1+z)^(1/z)]=e)
=1/e;
(2)原式=lim(x->0){[(1+2x)^(1/(2x))]^2}
={lim(x->0)[(1+2x)^(1/(2x))]}^2
=e^2 (应用重要极限lim(z->0)[(1+z)^(1/z)]=e)
=e²;
(3)原式=lim(n->∞){[(1+1/(n+1))^(1/(n+1))]^(2n/(n+1))}
={lim(n->∞)[(1+1/(n+1))^(1/(n+1))]}^[lim(n->∞)(2n/(n+1))]
=e^[lim(n->∞)(2/(1+1/n))] (应用重要极限lim(z->0)[(1+z)^(1/z)]=e)
=e^2
=e²。
(1)原式=e^{lim(x->0)[ln(1-x)/x]}
=e^{lim(x->0)[-1/(1-x)]} (0/0型极限,应用罗比达法则)
=e^(-1)
=1/e;
(2)原式=e^{lim(x->0)[ln(1+2x)/x]}
=e^{lim(x->0)[2/(1+2x)]} (0/0型极限,应用罗比达法则)
=e^2
=e²;
(3)原式=e^{lim(n->∞)[2nln((n+2)/(n+1))]}
=e^{lim(n->∞)[2ln((1+2/n)/(1+1/n))/(1/n)]}
=e^{lim(x->0)[2ln((1+2x)/(1+x))/x]} (令x=1/n)
=e^{lim(x->0)[2/((1+2x)(1+x))]} (0/0型极限,应用罗比达法则)
=e^2
=e²。
解法二:(重要极限法)
(1)原式=lim(x->0){[(1+(-x))^(1/(-x))]^(-1)}
={lim(x->0)[(1+(-x))^(1/(-x))]}^(-1)
=e^(-1) (应用重要极限lim(z->0)[(1+z)^(1/z)]=e)
=1/e;
(2)原式=lim(x->0){[(1+2x)^(1/(2x))]^2}
={lim(x->0)[(1+2x)^(1/(2x))]}^2
=e^2 (应用重要极限lim(z->0)[(1+z)^(1/z)]=e)
=e²;
(3)原式=lim(n->∞){[(1+1/(n+1))^(1/(n+1))]^(2n/(n+1))}
={lim(n->∞)[(1+1/(n+1))^(1/(n+1))]}^[lim(n->∞)(2n/(n+1))]
=e^[lim(n->∞)(2/(1+1/n))] (应用重要极限lim(z->0)[(1+z)^(1/z)]=e)
=e^2
=e²。
展开全部
你这题需要用到重要极限。1的无穷次幂极限为e。
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