Question

求极限x趋于0 时 （sinx/x）^(1/x^2)

Answer 1

x→0
lim (sinx/x)^(1/x^2)
=lim e^ln (sinx/x)^(1/x^2)
=e^lim ln (sinx/x)^(1/x^2)
考虑
lim ln (sinx/x)^(1/x^2)
=lim ln(sinx/x) / x^2
=lim ln(1+sinx/x - 1) / x^2
利用等价无穷小：ln(1+x)~x
=lim (sinx/x - 1) / x^2
=lim (sinx-x)/x^3
该极限为0/0型,利用L'Hospital法则
=lim (sinx-x)' / (x^3)'
=lim (cosx-1) / (3x^2)
该极限为0/0型,利用L'Hospital法则
=lim (cosx-1)' / (3x^2)'
=lim -sinx / 6x
根据重要的极限：lim sinx/x=1
=-1/6
因此,原极限=e^(-1/6)
有不懂欢迎追问