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let y=1/x
L =lim(x->∞) [ (1+1/x)^x /e ]^x
=lim(y->0) [ (1+y)^(1/y) /e ]^(1/y)
lnL
=lim(y->0) ln[ (1+y)^(1/y) /e ] /y (0/0)
=lim(y->0) { 1/[y(1+y)] -(1/y^2).ln(1+y) }
=lim(y->0) [y - (1+y)ln(1+y) ]/[y^2.(1+y)]
=lim(y->0) [y - (1+y).ln(1+y) ]/y^2 (0/0)
=lim(y->0) [1 - 1 - ln(1+y) ]/(2y)
=lim(y->0) -y/(2y)
=-1/2
=>
L = e^(-1/2)
L =lim(x->∞) [ (1+1/x)^x /e ]^x
=lim(y->0) [ (1+y)^(1/y) /e ]^(1/y)
lnL
=lim(y->0) ln[ (1+y)^(1/y) /e ] /y (0/0)
=lim(y->0) { 1/[y(1+y)] -(1/y^2).ln(1+y) }
=lim(y->0) [y - (1+y)ln(1+y) ]/[y^2.(1+y)]
=lim(y->0) [y - (1+y).ln(1+y) ]/y^2 (0/0)
=lim(y->0) [1 - 1 - ln(1+y) ]/(2y)
=lim(y->0) -y/(2y)
=-1/2
=>
L = e^(-1/2)
追问
老哥,我还是觉得你的解答靠谱些,可是最佳回答我给早了,实在不好意思。
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展开全部
lim(x→∞) (1+1/x)=e
所以原式=1
所以原式=1
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