大师作文网求极限（如图）,
来源：学生作业帮 编辑：大师作文网作业帮 分类：数学作业 时间：2024/11/12 06:50:09
求极限（如图）,
电脑输入不方便...方便起见,所有运算均是在x趋向于正无穷的情况下..实在不好打,见谅
(x-1)e^(π/2+arctanx) - xe^π
=e^π/2 * [(x-1) * e^arctanx - xe^π/2]
=e^π/2 * [x * (e^arctanx - e^π/2) -e^arctanx]
中括号里面的部分,最后面的e^arctanx当x趋向于正无穷为π/2；
前面那部分,作如下运算：
对 x * (e^arctanx - e^π/2)用罗比达法则,
x * (e^arctanx - e^π/2)
=[(e^arctanx - e^π/2)] / (1/x)
=[e^arctanx * 1/(1+x^2)] / (-1/x^2)
=-[x^2/(1+x^2)] * e^arctanx
=-1*π/2
=-π/2
带入最开始的式子,则
(x-1)e^(π/2+arctanx) - xe^π
=e^π/2 * [x * (e^arctanx - e^π/2) -e^arctanx]
=e^π/2 * (-e^π/2-e^π/2)
=-2e^π
输入格式限制,还请慢慢看..
(x-1)e^(π/2+arctanx) - xe^π
=e^π/2 * [(x-1) * e^arctanx - xe^π/2]
=e^π/2 * [x * (e^arctanx - e^π/2) -e^arctanx]
中括号里面的部分,最后面的e^arctanx当x趋向于正无穷为π/2；
前面那部分,作如下运算：
对 x * (e^arctanx - e^π/2)用罗比达法则,
x * (e^arctanx - e^π/2)
=[(e^arctanx - e^π/2)] / (1/x)
=[e^arctanx * 1/(1+x^2)] / (-1/x^2)
=-[x^2/(1+x^2)] * e^arctanx
=-1*π/2
=-π/2
带入最开始的式子,则
(x-1)e^(π/2+arctanx) - xe^π
=e^π/2 * [x * (e^arctanx - e^π/2) -e^arctanx]
=e^π/2 * (-e^π/2-e^π/2)
=-2e^π
输入格式限制,还请慢慢看..