x→0
lim (arctanx/x)^(1/x^2)
=lim e^ln (arctanx/x)^(1/x^2)
=e^lim ln (arctanx/x)^(1/x^2)
考虑
lim ln (arctanx/x)^(1/x^2)
=lim ln(arctanx/x) / x^2
=lim ln(1+arctanx/x-1) / x^2
根据等价无穷小：ln(1+x)~x
=lim (arctanx/x - 1) / x^2
=lim (arctanx-x) / x^3
该极限为0/0型，根据L'Hospital法则
=lim (arctanx-x)' / (x^3)'
=lim (1/(1+x^2) - 1) / 3x^2
=lim (1-1-x^2) / (1+x^2)(3x^2)
=lim -(x^2) / (3x^2)(1+x^2)
=lim -1 / 3(1+x^2)
=-1/3
故，原极限=e^(-1/3)
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