第1个回答 2015-11-10
可以直接用等价无穷小替换 极限值＝1 构成如下图：
第2个回答 2015-11-10
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)追问
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)追问
1为什么到下一步就没了