大学数学求极限,是高手的就进来看看! - 知识问答

大学数学求极限,是高手的就进来看看!

1个回答

log{[ntan(1/n)]^(n^2)}
=n^2*log[ntan(1/n)]
=n^2*log[n*(1/n+1/n^3/3)] (等价无穷小)
=n^2*log[1+1/n^2/3]
=n^2*1/n^2/3
=1/3
所以lim log...=1/3
lim... = e^(1/3)

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