计算下列极限：
(1)limx→0(tan5x/x)；(2)limlimx→0(sin2x/3x)；
(3)limx→0xcotx；(4)limx→0[(1-cos2x)/xsinx]；
(5)limn→∞2nsin(x/2n)；(6)limx→0(1+2x)1/x
(7)limx→0(1-3x)1/x；(8)limx→∞(1+1/x)3x；
(9)limx→∞(1-1/2x)x；(10)limn→∞[(2n+3)/(2n+1)]n+1
(1)limx→0(tan5x/x)=limx→0[(sin5x/5x)•(5/cos5x)]=[limx→0(sin5x/5x)]•[limx→0(5/cos5x)]=5 (2)limx→0(sin2x/3x)=limx→0(sin2x/2x•(2/3)=2/3 (3)limx→0xcotx=[limx→0(x/sinx)]•cosx=[limx→0(x/sinx)]•limx→0cosx=1 (4)limx→0[(1-cos2x)/xsinx]=limx→0[1-(1-2sin2x)]/xsinx=limx→0(2sinx/x)=2 (5)limn→∞2nsin(x/2n)=limn→∞[sin(x/2n)]/(x/2n)•x=limn→∞[sin(x/2n)]/(x/2n)•limn→∞x=x (6)limx→0(1+2x)1/x=limx→0[(1+2x)1/2x]2=e2 (7)limx→0(1-3x)1/x=limx→0(1-3x)-1/3x]=e-3 (8)limx→∞(1+1/x)3x=limx→∞(1+1/x)x]3=e3 (9)limx→∞(1-1/2x)x=limx→∞(1-1/2x)-2x]-(1/2)=e-(1/2) (10)limn→∞[(2n+3)/(2n+1)]n+1=limn→∞[1+2/(2n+1)(2n+1)/2)]2/(2n+1)•(n+1)
