求下面两个极限

lim x[(x^2+1)^(1/2)-x]=lim x/ [(x^2+1)^(1/2)+x]
用洛必达法则=lim 1/[x/(x^2+1)^(1/2)+1]=
lim (x^2+1)^(1/2)/[(x^2+1)^(1/2)+x]=1-limx/ [(x^2+1)^(1/2)+x]
得到2lim x/ [(x^2+1)^(1/2)+x]=1
即 lim x/ [(x^2+1)^(1/2)+x]=1/2
lim x[(x^2+1)^(1/2)-x]=1/2

lim (tanx-sinx)/x^3=lim sinx(1-cosx)/(x^3*cosx)
=lim sinx*2[sin(x/2)]^2/(x^3*cosx)=
lim sinx/x*lim (sin(x/2))^2/(x/2)^2*lim1/2cosx=1/2

第一个1/2

limx[(x^2+1)^(1/2)-x]
＝lim[x／（√（x²+1）＋x）]
＝lim[1／（√（1+1/x²）＋1]
＝1/2.

lim (tanx-sinx)/x³＝lim[sinx（1-cosx）]／x³cosx
＝lim（1-cosx）／x²＝1/2
（sinx∽x,（1-cosx）∽x²/2）

第一题
原式=lim x[(x^2+1)^(1/2)-x)*[(x^2+1)^(1/2)+x]/[(x^2+1)^(1/2)+x] x趋近于+∞
=lim x/[(x^2+1)^(1/2)+x] x趋近于+∞
上下再同时出x得：
=lim 1/{[(x^2+1)^(1/2)+x]/x+1} x趋近于+∞
= 1/2

第二题
=lim sinx(1-cosx)/(x^3*cosx) x趋近于0
=lim sinx*2[sin(x/2)]^2/(x^3*cosx) x趋近于0
=lim sinx/x*lim (sin(x/2))^2/(x/2)