f(1)=0,f(x)在1点处的导数存在,求[f(sin^2x+cosx)]/[(e^x-1)tgx](x趋于0)的值.

Question

f(1)=0,f(x)在1点处的导数存在,求[f(sin^2x+cosx)]/[(e^x-1)tgx](x趋于0)的值.

Answer 1

LIM(e^x-1)/(sinxcosx)
=LIM(e^x-1)/sinx *cos0
=LIM(e^x/cosx)
=1
LIM[f(sin^2x+cosx)]/[(e^x-1)tgx](x趋于0)
=LIM[f'(sin^2x+cosx)(2sinxcosx-sinx)/(e^x *tgx+(e^x-1)*sec^2x)(x趋于0)
=LIMf'(sin^2x+cosx)*LIM(2sinxcosx-sinx)/(e^x*tgx+(e^x-1)sec^x)
=f'(1)*LIM(2cos^2x-cosx)/(e^x+(e^x-1)/sinxcosx)
=f'(1)*(2-1)/(1+1)
=f'(1)/2