如何用洛必达法则求lim x→0+ (sinx)^(k/1+ln x) (k为常数)
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请问你的指数部分是什么,k/(1+lnx)?
取自然对数lim (x→0+)ln (sinx)^(k/1+ln x)
=lim (x→0+)ln (sinx)*k/(1+ln x) (0/0,用洛必达法则)
=lim (x→0+)cosx/sinx*k/(1/ x)
=lim (x→0+)kcosx*x/sinx
=k因此lim (x→0+) (sinx)^(k/1+ln x)
=lim (x→0+)e^ln (sinx)^(k/1+ln x)
=e^k