求高等数学极限问题。谢谢大家了!!
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罗比达法则:
先求y=(1+1/n)^2n的导数:
lny=2nln(1+1/n)
y'/y=2ln(1+1/n)+2n/(1+1/n)*(-1/n^2)
y'/y=2ln(1+1/n)-2/(n+1)
y'=[2ln(1+1/n)-2/(n+1)]*(1+1/n)^2n
lim n(e^2-(1+1/n)^2n)
=lim (e^2-(1+1/n)^2n)/(1/n)
=lim -{[2ln(1+1/n)-2/(n+1)]*(1+1/n)^2n}/(-1/n^2)
lim [ln(1+1/n)-1/(n+1)]/(1/n^2)
=lim [1/(1+1/n)*(-1/n^2)+1/(n+1)^2]/(-2/n^3)
=lim n^2/2(n+1)^2
=1/2.
所以:
lim n(e^2-(1+1/n)^2n)
=2*e^2*1/2
先求y=(1+1/n)^2n的导数:
lny=2nln(1+1/n)
y'/y=2ln(1+1/n)+2n/(1+1/n)*(-1/n^2)
y'/y=2ln(1+1/n)-2/(n+1)
y'=[2ln(1+1/n)-2/(n+1)]*(1+1/n)^2n
lim n(e^2-(1+1/n)^2n)
=lim (e^2-(1+1/n)^2n)/(1/n)
=lim -{[2ln(1+1/n)-2/(n+1)]*(1+1/n)^2n}/(-1/n^2)
lim [ln(1+1/n)-1/(n+1)]/(1/n^2)
=lim [1/(1+1/n)*(-1/n^2)+1/(n+1)^2]/(-2/n^3)
=lim n^2/2(n+1)^2
=1/2.
所以:
lim n(e^2-(1+1/n)^2n)
=2*e^2*1/2
追问
你这是哪道题目的答案?不是我问的啊。
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