求当x→0时，（arctanx ⼀ x）^ (1 ⼀ (x^2) )的极限，答案是e^(-1⼀3).求较为详细的过程，谢谢！

（arctanx / x）^ (1 / (x^2) ) = exp(Ln(arctanx / x) / x^2)
直接求e的指数的极限 Ln(arctanx / x) / x^2
Ln(arctanx /x -1 + 1) / x^2
= arctanx/x -1 / x^2
(arctanx - x) / x^3
罗比达法则
= [(1/1+x^2) - 1]/3x^2
= - 1/3(1+x^2)
极限是-1/3
在带回 e 极限是 e^(-1/3)

这是个1^∞极限
先取自然对数
lim(x→0)ln（arctanx / x）^ (1 / (x^2) )
=lim(x→0)ln（arctanx / x）/x^2
=lim(x→0)[ln（arctanx )-ln x]/x^2 (0/0型，上下求导）
=lim(x→0)[1/arctanx *1/(1+x^2)-1/x]/(2x)
=lim(x→0)[x-arctanx*(1+x^2)]/[2x^2arctanx (1+x^2)]
=lim(x→0)[x-arctanx*(1+x^2)]/[2x^3] (0/0型，上下求导)
=lim(x→0)[1-1-2xarctanx]/[6x^2]
=lim(x→0)[-2xarctanx]/[6x^2]
=-1/3
所以
lim(x→0)（arctanx / x）^ (1 / (x^2) )
=lim(x→0)e^ln（arctanx / x）^ (1 / (x^2) )
=e^(-1/3)

y=（arctanx / x）^ (1 / (x^2) )
lny=ln(arctanx/x)/x^2
x->0,0/0,洛必达
=[ln(arctanx)- lnx]/x^2
=[1/arctanx*1/(1+x^2)-1/x]/2x
=[x-(1+x^2)arctan x]/2x^2(1+x^2)arctanx
0/0
洛必达
=[1-2xarctanx-1]/[(4x+8x^3)arctanx+2x^2]
=-arctanx/[(2+4x^2)arctanx+x]
0/0
洛必达
=-[1/(1+x^2)]/[8xarctanx+(2+4x^2)/(1+x^2)+1]
把x=0代入
=-[1/(1+0)]/[0+(2+0)/(1+0)+1]=-1/3
所以y->e^(-1/3)