高等数学吧 关注：472,031贴子：3,874,212
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塔纳托斯
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x->0
(1+x)^(1/x)
=e^[ln(1+x)/x]
=e^{[x-(1/2)x^2+(1/3)x^3+o(x^3)]/x}
=e^[1-(1/2)x+(1/3)x^2+o(x^2)]
=e. e^[-(1/2)x+(1/3)x^2+o(x^2)]
=e. { 1+[-(1/2)x+(1/3)x^2] +(1/2)[-(1/2)x+(1/3)x^2]^2 +o(x^2)] }
=e. { 1+[-(1/2)x+(1/3)x^2] +(1/2)[(1/4)x^2+...] +o(x^2)] }
=e. [1-(1/2)x+(11/24)x^2+o(x^2)]
(1+x)^(1/x) -e +(e/2)x =e. [(11/24)x^2+o(x^2)]
lim(x->0) [(1+x)^(1/x) -e +(e/2)x]/x^2
=lim(x->0) e. [(11/24)x^2] /x^2
=(11/24)e
(1+x)^(1/x)
=e^[ln(1+x)/x]
=e^{[x-(1/2)x^2+(1/3)x^3+o(x^3)]/x}
=e^[1-(1/2)x+(1/3)x^2+o(x^2)]
=e. e^[-(1/2)x+(1/3)x^2+o(x^2)]
=e. { 1+[-(1/2)x+(1/3)x^2] +(1/2)[-(1/2)x+(1/3)x^2]^2 +o(x^2)] }
=e. { 1+[-(1/2)x+(1/3)x^2] +(1/2)[(1/4)x^2+...] +o(x^2)] }
=e. [1-(1/2)x+(11/24)x^2+o(x^2)]
(1+x)^(1/x) -e +(e/2)x =e. [(11/24)x^2+o(x^2)]
lim(x->0) [(1+x)^(1/x) -e +(e/2)x]/x^2
=lim(x->0) e. [(11/24)x^2] /x^2
=(11/24)e
