Question

一道高数极限题求助

Answer 1

这道题一定要注意反三角函数的取值范围，也就是要注意x趋紧0正或0负的情况，要分类讨论，同时要运用等价无穷小的公式，具体步骤如下：

希望能帮到你

Answer 2

lim(x->0+) (e^x -1) arctan(1/x) /√[ cosx - (cosx)^2]
=(π/2) lim(x->0+) (e^x -1) /√[ cosx - (cosx)^2]
=(π/2) lim(x->0+) (e^x -1) / [√cosx .√( 1-cosx) ]
=(π/2) lim(x->0+) (e^x -1) / √( 1-cosx)
=(π/2) lim(x->0+) x/ √( 1-cosx)
=(π/2) lim(x->0+) x/ [(√2/2)x ]
=(√2/2)π
lim(x->0-) (e^x -1) arctan(1/x) /√[ cosx - (cosx)^2]
=-(π/2) lim(x->0-) (e^x -1) /√[ cosx - (cosx)^2]
=-(π/2) lim(x->0-) (e^x -1) / [√cosx .√( 1-cosx) ]
=-(π/2) lim(x->0-) (e^x -1) / √( 1-cosx)
=-(π/2) lim(x->0-) x/ √( 1-cosx)
=-(π/2) lim(x->0-) x/ [-(√2/2)x ]
=(√2/2)π
=>
lim(x->0) (e^x -1) arctan(1/x) /√[ cosx - (cosx)^2]
=(√2/2)π