Question

一道有关微积分题目求大佬解答

Answer 1

x->0

(1+x)^(1/x)

=e^[ln(1+x)/x]

ln(1+x) = x -(1/2)x^2 +(1/3)x^3 +o(x^3)

=e^[ 1 - (1/2)x + (1/3)x^2 +o(x^2)]

=e. e^[ - (1/2)x + (1/3)x^2 +o(x^2)]

利用泰勒公式把 e^[ - (1/2)x + (1/3)x^2 +o(x^2)] 拆开

=e. { 1 +[ - (1/2)x + (1/3)x^2 +o(x^2)] +(1/2)[ - (1/2)x + (1/3)x^2 +o(x^2)]^2 +o(x^2) }

=e. { 1 +[ - (1/2)x + (1/3)x^2 +o(x^2)] +(1/2)[ (1/4)x^2 +o(x^2)] +o(x^2) }

=e. [1 -(1/2)x + (11/24)x^2 +o(x^2) ]

e^[(1+x)^(1/x)]

=e^【e. [1 -(1/2)x + (11/24)x^2 +o(x^2) ]】

=e^e. e^[ -(e/2)x + (11e/24)x^2 +o(x^2) ]

利用泰勒公式把 e^[ -(e/2)x + (11e/24)x^2 +o(x^2) ] 拆开

=e^e. { 1+[-(e/2)x + (11e/24)x^2+o(x^2)]+(1/2)[-(e/2)x+(11e/24)x^2 +o(x^2)]^2 }

=e^e. { 1+[-(e/2)x + (11e/24)x^2+o(x^2)]+(1/2)[(e^2/4)x^2+o(x^2)] }

=e^e. { 1 -(e/2)x + [(11e/24) + (e^2/8) ]x^2+o(x^2) } 

同样地

(1+x)^(e/x)

=e^[e.ln(1+x)/x]

=e^[ e -(e/2)x + (e/3)x^2 +o(x^2)]

=e^e. e^[-(e/2)x + (e/3)x^2 +o(x^2)]

=e^e.{ 1 +[-(e/2)x + (e/3)x^2 +o(x^2)] +(1/2)[-(e/2)x + (e/3)x^2 +o(x^2)]^2 +o(x^2) }

=e^e.{ 1 +[-(e/2)x + (e/3)x^2 +o(x^2)] +(1/2)[(e^2/4)x^2 +o(x^2)] +o(x^2) }

=e^e.{ 1 -(e/2)x + [(e/3)+(e^2/8)]x^2 +o(x^2) }

e^[(1+x)^(1/x)] -(1+x)^(e/x)

=e^e. { 1 -(e/2)x + [(11e/24) + (e^2/8) ]x^2+o(x^2) } 

-e^e.{ 1 -(e/2)x + [(e/3)+(e^2/8)]x^2 +o(x^2) }

=e^e.   [(e/8)x^2 +o(x^2) ]

lim(x->0) { e^[(1+x)^(1/x)] -(1+x)^(e/x) }/x^2

=lim(x->0) e^e.   [(e/8)x^2 ]/x^2

=(e/8) e^e

=(1/8) e^(e+1)