计算下列极限：
(1)limx→0(ex-e-x)/sinx
(2)limx→π(sin2x/tan5x)；
(3)limx→+∞；
(4)limx→π(π-x)tan(x/2)；
(5)limx→0[1/x-1/(ex-1)]；
(6)limx→1x1/(x-1)；
(7)limx→+∞(lnx)1/x；
(8)limx→0+(tanx)x
(9)limx→0-lnx•ln(1-x)；
(10)limx→0((2/π)arccosxx)1/x；
(11)limx→∞(31/n-1)
(1)limx→0[(ex-e-x)/sinx]=limx→0[(ex+e-x)/cosx]=2 (2)limx→π(sin2x/tan5x)=limx→π(2cos2x/5sec25x)=2/5 (3)limx→+∞[ln(1+1/x)/arcosx]=limx→+∞[1/(1+1/x2)•(-(1/x2))/-[1/(1+x2)]=limx→+∞[(1+x2)/(x2+x)]=1 (4)limx→π(π-x)tan(x/2)=limx→π(π-x)/[1/tan(x/2)]=limlimx→π-1/{[-sec2(x/2)•(1/2)]/[tan2(x/2)]} =limx→π{2tan2(x/2)/[sec2(x/2)]}=limx→π2sin2(x/2)=2 (5)limx→0[1/x-(1/ex-1)=limx→0{(ex-1-x)/[x(ex-1)]}=limx→0[(ex-1)/ex-1+xex] =limx→0[ex/ex+ex+xex]=limx→0[1/(2+x)]=1/2 (6)limx→11/(x-1)=limx→1elnx/(x-1)=elimx→1lnx/(x-1)=elimx→11/x=e (7)limx→+∞(lnx)1/x=limx→+∞elnx/x=elimx→+∞lnx/x=elimx→+∞1/x=e0=1 (8)limx→0+(tanx)x=limx→0++elntanx/(1/x)=elimx→0+[lntanx/(1/x)]=elimx→0+[1/tanx•(1/cos2x)]/-(1/x2) =elimx→0+[x/sinx•(x/cosx)]=e0=1 (9)limx→1-lnxln(1-x)=limx→1-[ln(1-x)/(1/lnx)] =limx→1-[-1/(1-x)]/[-(1/x)/ln2x]=limx→1-[xln2x/(1-x)] =limx→1-[ln2x+x•2lnx•(1/x)]/-1=0 (10)limx→0[(2/π)arccosx]1/x=limx→0ln(2/π)arccosx/x=elimx→0ln(2/π)arccosx/x =elimx→0-1/√(1-x2)/arccosx=e-2/π (11)因为 limx→∞x(31/x-1)=limx→∞[(31/x-1)/(1/x)] =limx→∞[31/xln3•(-1/x2)]/-(1/x2) =limx→∞31/xln3=ln3 所以 limn→∞n(31/x-1)=ln3
