极限求解lim[(2/π)arctanx]^x (x趋向于正的无穷大）

x→+∞
lim [(2/π)arctanx]^x
=lim e^ln [(2/π)arctanx]^x
=e^lim ln[(2/π)arctanx]^x
考虑
lim ln[(2/π)arctanx]^x
=lim x * ln[(2/π)arctanx]
=lim ln[1+(2/π)arctanx-1] / (1/x)
=lim [(2/π)arctanx-1] / (1/x)
该极限为0/0型,根据L'Hospital法则
=lim [(2/π)arctanx-1]' / (1/x)'
=lim (2/π)/(x^2+1) / (-1/x^2)
=(-2/π)*lim (x^2)/(1+x^2)
=-2/π
因此,原极限=e^(-2/π)
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