求下列极限
(1)limx→0(x-sinx)/(x+x2)
(2)limx→+∞{[(1+1/x)]cos(1/x)}/arccotx
(3)limx→+∞ex/xx
(4)limx→0-lncotx/lnx
(5)limx→+∞(ex-ex)/(ex+ex)
(6)lim[ln(1+sin2x)/arcsin(x+x2)]
(7)limx→0(x-sinx)/x2ln(1+x)
(8)limx→2+[cosx•ln(x-2)/ln(ex-e2)]
(9)limx→0(1-sinx)x/2
(10)limx→0(sinx/x)x/1
(11)limx→1[x/(x-1)-1/lnx]
(12)limx→0+xlnsinx
【正确答案】：(1)limx→0(x-sinx)/(x+x3)=x→0(1-cosx)/(1+3x2)=0 (2)limx→+∞{[ln(1+1/x)]cos(1/x)}/arccotx =limx→+∞[ln(1/x)/arccotx]limx→+∞cos(1/x) =limx→+∞{1/(1+1/x)•(-1/x2)/[-1(1+x2)]}•1 =limx→+∞(1+x2)/[x(1+x)]=1 (3)limx→+∞ex/xn=limx→+∞(ex/nxn-1) =limx→+∞ex/n(n-1)xn-2 =…=limx→+∞ex/n!=+∞ (4)limx→0+lncotx/lnx=limx→0+tanx•(-csc2x)/(1/x) =limx→0+xsinx/cosxsin2x=1 (5)limx→+∞(ex-e-x)/(ex+e-x)=limx→+∞(ex+e-x)/(ex-e-x) =limx→+∞(ex-e-x)/(ex+e-x)继续下去将一直循环出现，换另法． limx→+∞(ex-e-x)/(ex+e-x)=1imx→+∞(1-e-2x)/(1+e-2x)=1 (6)limx→0ln(1+sin2x)/arcsin(x+x2)=limx→0sin2x/(x+x2) =limx→02x/(x+x2)=2 (7)limx→0(x-sinx)/x2ln(1+x)=limx→0(x-sinx)/x3 =limx→0(1-cosx)/3x2 =limx→0sinx/6x=1/6． (8) limx→2+[cosx•ln(x-2)]/ln(ex-e2)=cos2•limx→2+[1/(x-2)]/[ex/(ex-e2)] =cos2•limx→2+(ex-e2)/(x-2)ex =cos2/e2llmx→2+(ex/1)=cos2． (9) limx→0(1-sinx)2/x=limx→0/sub>[1+(-sinx)]1/-sinx•-2sinx/x =e-2 (10) limx→0(sinx/x)1/x=limx→0e(1/x)ln(sinx/x)=elimx→0ln(sinx/x)/x limx→0ln(sinx/x)/x=limx→0(xcosx-sinx)/xsinx =limx→0(xcosx-sinx)/x2 =limx→0-xsinx/2x=0． 所以，原式=e0=1 (11)limx→1[x/(x-1)-1/lnx]limx→1(xlnx-x+1)/(x-1)lnx=limx→1xlnx/(xlnx+x-1) =limx→1(lnx+1)/(lnx+2)=1/2 (12)limx→0+xlnsinx=limx→0+lnsinx/x-1 =limx→0+-x2cosx/sinx =limx→0+-x2cosx/x=0
