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limxx 1 1lnx

x/(x-1)-1/lnx=(xlnx-x+1)/(x-1)lnx。当x趋于1时,分子分母都趋于0。故可用上下求导:即:lim(xlnx-x+1)/(x-1)lnx=lim(lnx+1-1)/(lnx+(x-1)/x)=lim(xlnx)/(xlnx+x-1)=lim(lnx+1)/(lnx+2)=1/2。即极限为1/2。

lim(x→1)[1/lnx-1/(x-1)] =lim(x→1)[x-1-lnx]/[lnx(x-1)](这是0/0型,运用洛必达) =lim(x→1)(1-1/x)/[(x-1)/x+lnx] =lim(x→1)(x-1)/(x-1+xlnx)(再运用洛必达法则) =lim(x→1)1/(1+lnx+1) =1/2

运用极限的四则运算法则,和洛必达法则求解 lim(x->+∞) 1/(x-lnx) =lim(x->+∞) (1/x)/(1-lnx/x) =lim(x->+∞) (1/x)/[1-lim(x->+∞) (lnx/x)] =0/[1-lim(x->+∞) (1/x)] =0/(1-0) =0

lim(x→1)x-x^x/1-x+lnx 这是0/0型,用洛必达法则,原式=lim(x→1)[1-(1+lnx)x^x]/(1-x)继续用洛必达,原式=lim(x→1)[x^(x-1)+(1+lnx)^2x^x]=1+1=2

原式=lim(x→+∞)xln(1+1/x) 设x=1/t,所以t→0 原式=lim(t→0)ln(1+t)/t=1

当x→0时,ex-1~x,所以当x→1时,x-xx=x(1-xx-1)=-x[e(x-1)lnx-1]~-(x-1)lnx.又因为当x→0时,ln(1+x)=x+12x2+o(x2),所以,当x→1时,ln(x)=ln(1+(x-1))=x-1+12(x?1)2+o((x-1)2).综上可得,limx→1x?xx1?x?lnx=limx→1?(x?1...

lim[x^(1/x)-1]^(1/lnx)=e^limln[x^(1/x)-1]/lnx罗比达法则=e^lim[1/(x^1/x -1)*(x^1/x)']/(1/x)因为x^1/x 化为自然对数求导后可得x^1/x=x^1/x *(1-lnx)/ x^2又因为limx趋向无穷大时x^1/x -1 =e^lnx/x -1 ~lnx/x 也可得limx^1/x =1代入=e^li...

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