求这个极限的过程,答案是二分之e ,谢谢了(^~^)
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(1+x)^(1/x)
= e^[ ln(1+x)/x]
= e^{ [ x -(1/2)x^2 ] /x} +o(x)
=e^[ 1- (1/2)x ] +o(x)
= e. e^(-1/2)x + o(x)
= e. [ 1- (1/2)x] + o(x)
Similary
(1+2x)^[1/(2x)]
= e^[ ln(1+2x)/(2x)]
= e^{ [ 2x -(1/2)(2x)^2 ] /(2x) } +o(x)
=e^( 1- x ) +o(x)
= e. e^(-x) + o(x)
= e. [ 1- x] + o(x)
(1+x)^(1/x) -(1+2x)^[1/(2x)]
= e. [ 1- (1/2)x] -e. [ 1- x] + o(x)
=[(1/2)e ]x + o(x)
lim(x->0) { (1+x)^(1/x) -(1+2x)^[1/(2x)] }/ sinx
=lim(x->0) [(1/2)e ]x / x
=(1/2)e
= e^[ ln(1+x)/x]
= e^{ [ x -(1/2)x^2 ] /x} +o(x)
=e^[ 1- (1/2)x ] +o(x)
= e. e^(-1/2)x + o(x)
= e. [ 1- (1/2)x] + o(x)
Similary
(1+2x)^[1/(2x)]
= e^[ ln(1+2x)/(2x)]
= e^{ [ 2x -(1/2)(2x)^2 ] /(2x) } +o(x)
=e^( 1- x ) +o(x)
= e. e^(-x) + o(x)
= e. [ 1- x] + o(x)
(1+x)^(1/x) -(1+2x)^[1/(2x)]
= e. [ 1- (1/2)x] -e. [ 1- x] + o(x)
=[(1/2)e ]x + o(x)
lim(x->0) { (1+x)^(1/x) -(1+2x)^[1/(2x)] }/ sinx
=lim(x->0) [(1/2)e ]x / x
=(1/2)e
追问
谢谢😜
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