x趋于0时求lim(sin/x)^(1/[1-cosx])
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x趋于0时求lim(sin/x)^(1/[1-cosx])
是sinx/x
是sinx/x
先用L'Hospital法则计算
lim(x→0)[1/(1-cosx)]*ln(sinx/x)
= lim(x→0)ln(sinx/x)/(1-cosx) (0/0)
= lim(x→0)[1/(sinx/x)]*[(xcosx-sinx)/x^2]/sinx
= lim(x→0)(xcosx-sinx)/x(sinx)^2
= lim(x→0)(xcosx-sinx)/x^3 (0/0)
= lim(x→0)(-xsinx)/3x^2
= -1/3,
于是,
lim(sinx/x)^(1/[1-cosx])
= e^(-1/3).
lim(x→0)[1/(1-cosx)]*ln(sinx/x)
= lim(x→0)ln(sinx/x)/(1-cosx) (0/0)
= lim(x→0)[1/(sinx/x)]*[(xcosx-sinx)/x^2]/sinx
= lim(x→0)(xcosx-sinx)/x(sinx)^2
= lim(x→0)(xcosx-sinx)/x^3 (0/0)
= lim(x→0)(-xsinx)/3x^2
= -1/3,
于是,
lim(sinx/x)^(1/[1-cosx])
= e^(-1/3).
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