第1个回答 2020-08-30
ln(sinx/x)^1/(x^2)＝ln(sinx/x)/(x^2)
x→0时,ln(sinx/x)＝ln[1＋(sinx－x)/x]等价于(sinx－x)/x,所以
lim(x→0) ln(sinx/x)^1/(x^2)＝lim(x→0) (sinx－x)/x^3＝lim(x→0) (cosx－1)/(3x^2)＝lim(x→0) (－1/2×x^2)/(3x^2)＝－1/6
所以,lim(x→0) (sinx/x)^1/(x^2)＝e^(-1/6)
x→0时,ln(sinx/x)＝ln[1＋(sinx－x)/x]等价于(sinx－x)/x,所以
lim(x→0) ln(sinx/x)^1/(x^2)＝lim(x→0) (sinx－x)/x^3＝lim(x→0) (cosx－1)/(3x^2)＝lim(x→0) (－1/2×x^2)/(3x^2)＝－1/6
所以,lim(x→0) (sinx/x)^1/(x^2)＝e^(-1/6)
第2个回答 2021-11-01
简单计算一下即可，答案如图所示
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