Question

求极限，答案是e，想知道过程

答案是e，怎么计算啊，希望有过程

Answer 1

令 y = [1+e^(1/x)]^ln(1+x),
则 lny = ln(1+x)ln[1+e^(1/x)] = ln[1+e^(1/x)]/[1/ln(1+x)]
lim<x→0+>lny = lim<x→0+>ln[1+e^(1/x)]/[1/ln(1+x)] (∞/∞)
= lim<x→0+>{(-1/x^2)e^(1/x)/[1+e^(1/x)]}/{[-1/(1+x)]/[ln(1+x)]^2}
= lim<x→0+>(1+x)e^(1/x)[ln(1+x)]^2/{x^2[1+e^(1/x)]}
= lim<x→0+>x^2e^(1/x)/{x^2[1+e^(1/x)]}
= lim<x→0+>e^(1/x)/[1+e^(1/x)] = lim<x→0+>1/[e^(-1/x)+1] = 1,
则 lim<x→0+> y = e

Answer 2

L=lim(x->0+) [ 1+ e^(1/x) ]^[ln(1+x)]

lnL
=lim(x->0+) ln[ 1+ e^(1/x) ]. ln(1+x)
=lim(x->0+) ln[ 1+ e^(1/x) ] /[1/ln(1+x)]
洛必达

=lim(x->0+) { -(1/x^2).e^(1/x)/[ 1+ e^(1/x) ] } / { [-1/(1+x)]/[ln(1+x)]^2 }
=lim(x->0+) e^(1/x). [ln(1+x)]^2 .(1+x)/ { x^2.[ 1+ e^(1/x) ] }
=lim(x->0+) e^(1/x). [ln(1+x)]^2 / { x^2.[ 1+ e^(1/x) ]}
=lim(x->0+) e^(1/x). (x^2) / { x^2.[ 1+ e^(1/x) ]}
=lim(x->0+) e^(1/x) / [ 1+ e^(1/x) ]
分子分母同时除 e^(1/x)
=lim(x->0+) 1 / [ 1/e^(1/x)+ 1 ]
=1
=>
L =e

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