图中的题怎么做呢
展开全部
lim [(x²-1)/(x-1)]·e^[1/(x-1)]
x→1⁺
=lim [(x+1)(x-1)/(x-1)]·e^[1/(x-1)]
x→1⁺
=lim (x+1)·e^[1/(x-1)]
x→1⁺
=2·e^(+∞)
=+∞
lim [(x²-1)/(x-1)]·e^[1/(x-1)]
x→1⁻
=lim [(x+1)(x-1)/(x-1)]·e^[1/(x-1)]
x→1⁻
=lim (x+1)·e^[1/(x-1)]
x→1⁻
=2·e⁰
=2
[(x²-1)/(x-1)]·e^[1/(x-1)]在x=1处两侧极限不相等
lim [(x²-1)/(x-1)]·e^[1/(x-1)] 不存在
x→1
x→1⁺
=lim [(x+1)(x-1)/(x-1)]·e^[1/(x-1)]
x→1⁺
=lim (x+1)·e^[1/(x-1)]
x→1⁺
=2·e^(+∞)
=+∞
lim [(x²-1)/(x-1)]·e^[1/(x-1)]
x→1⁻
=lim [(x+1)(x-1)/(x-1)]·e^[1/(x-1)]
x→1⁻
=lim (x+1)·e^[1/(x-1)]
x→1⁻
=2·e⁰
=2
[(x²-1)/(x-1)]·e^[1/(x-1)]在x=1处两侧极限不相等
lim [(x²-1)/(x-1)]·e^[1/(x-1)] 不存在
x→1
展开全部
上面不是写的D吗?
追问
不会算
追答
lim [(x²-1)/(x-1)]·e^[1/(x-1)]
x→1⁺
=lim [(x+1)(x-1)/(x-1)]·e^[1/(x-1)]
x→1⁺
=lim (x+1)·e^[1/(x-1)]
x→1⁺
=2·e^(+∞)
=+∞
lim [(x²-1)/(x-1)]·e^[1/(x-1)]
x→1⁻
=lim [(x+1)(x-1)/(x-1)]·e^[1/(x-1)]
x→1⁻
=lim (x+1)·e^[1/(x-1)]
x→1⁻
=2·e⁰
=2
[(x²-1)/(x-1)]·e^[1/(x-1)]在x=1处两侧极限不相等
lim [(x²-1)/(x-1)]·e^[1/(x-1)] 不存在
x→1
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