求limx^tanx在x趋向于0+时的极限.用洛必达法则.

Question

求limx^tanx在x趋向于0+时的极限.用洛必达法则.

Answer 1

令y=x^tanx,且两边同时取对数.则lim(x趋于0+)x^tanx=lim(x趋于0+)y=lim(x趋于0+)e^lny=lim(x趋于0+)e^(tanx*lnx)=e^[lim(x趋于0+)tanxlnx].现在分析[lim(x趋于0+)tanxlnx],=lim(x趋于0+) [ lnx / (1/tanx) ]=lim(x趋于0+) (1/x) / { [-sec^(2)x]/[tan^(2)x] }=lim(x趋于0+) (1/x)/[(-secx/tanx)^2] =lim(x趋于0+) (1/x) / -{ [1/cosx)/tanx]^2 }=lim(x趋于0+) (1/x) / -{1/[sin^(2)x] }=lim(x趋于0+) -sin^(2)x/x=lim(x趋于0+) (x^2)/x=-x=0,所以待求极限为e^0=1.