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(sinx/x)^x -1
=e^[x.ln(sinx/x)] -1
/
x->0
sinx ~ x -(1/6)x^3
sinx/x ~ 1- (1/6)x^2
ln(sinx/x) ~ln[1- (1/6)x^2] ~ -(1/6)x^2
x.ln(sinx/x) ~ -(1/6)x^3
e^[x.ln(sinx/x)] ~ e^[(-1/6)x^3] ~ 1-(1/6)x^3
e^[x.ln(sinx/x)]-1 ~ -(1/6)x^3
x^x . [(sinx/x)^x -1] ~x^x.[-(1/6)x^3] ~ -(1/6)x^(x+3)
x^x . [(sinx/x)^x -1] /x^3 ~-(1/6)x^(x+3)/x^3 ~ -(1/6)x^x
/
-lim(x->0+) x^x.[ (sinx/x)^x -1 ]
=-lim(x->0+) -(1/6)x^x
=1/6
=e^[x.ln(sinx/x)] -1
/
x->0
sinx ~ x -(1/6)x^3
sinx/x ~ 1- (1/6)x^2
ln(sinx/x) ~ln[1- (1/6)x^2] ~ -(1/6)x^2
x.ln(sinx/x) ~ -(1/6)x^3
e^[x.ln(sinx/x)] ~ e^[(-1/6)x^3] ~ 1-(1/6)x^3
e^[x.ln(sinx/x)]-1 ~ -(1/6)x^3
x^x . [(sinx/x)^x -1] ~x^x.[-(1/6)x^3] ~ -(1/6)x^(x+3)
x^x . [(sinx/x)^x -1] /x^3 ~-(1/6)x^(x+3)/x^3 ~ -(1/6)x^x
/
-lim(x->0+) x^x.[ (sinx/x)^x -1 ]
=-lim(x->0+) -(1/6)x^x
=1/6
追问
思路好清晰。感谢
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