n趋于无穷,n^2 / (e的根号下n次方)的极限怎么算?
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n趋于无穷,n^2 / (e的根号下n次方)的极限怎么算?
![n趋于无穷,n^2 / (e的根号下n次方)的极限怎么算?](/uploads/image/z/17157997-37-7.jpg?t=n%E8%B6%8B%E4%BA%8E%E6%97%A0%E7%A9%B7%2Cn%5E2+%2F+%28e%E7%9A%84%E6%A0%B9%E5%8F%B7%E4%B8%8Bn%E6%AC%A1%E6%96%B9%29%E7%9A%84%E6%9E%81%E9%99%90%E6%80%8E%E4%B9%88%E7%AE%97%3F)
用洛必达法则
n^2/e^√n
上下求导
=2n/[e^√n*(1/2√n)]
=4n√n/e^√n
=4n^(3/2)/e^√n
再求导
=6n^(1/2)/[e^√n*(1/2√n)]
=12n/e^√n
再求导
=12/[e^√n*(1/2√n)]
=24√n/2/e^√n
再求导
=24*(√n)'/[e^√n*(√n)']
=24/e^√n
n趋于无穷,所以极限等于0
n^2/e^√n
上下求导
=2n/[e^√n*(1/2√n)]
=4n√n/e^√n
=4n^(3/2)/e^√n
再求导
=6n^(1/2)/[e^√n*(1/2√n)]
=12n/e^√n
再求导
=12/[e^√n*(1/2√n)]
=24√n/2/e^√n
再求导
=24*(√n)'/[e^√n*(√n)']
=24/e^√n
n趋于无穷,所以极限等于0
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