lim[1÷(sinx)^2 - (cosx)^2÷x^2],求当x→0是的极限,

lim (e^x-sinx-1)/(arcsinx^2) lim(x+e^2x)^(1/sinx) 请问高手 x趋近于0 lim (sinx-x*cosx)/(sinx)^3 能不能这样计算原式=lim sinx/(sinx)^3-lim (x*cosx)/(sinx)^3=lim 1/(sinx)^2-lim cosx/(sinx)^2=lim (1-cosx)/(sinx)^2=lim (x^2/2)/(x^2)=1/2,如果可以为何与利用麦克劳林公式所得 lim(1-sinx)^1/x lim [xtan(2/x)+(1/x^2)sinx^2] . lim（2x+1）sinx/2/x+2= lim 1/(sinx)^2-1/xsinx,x趋近无穷大 lim(x→0)(1+sinx)^(1/2x) lim((e^2-1-x)^2)/(x(sinx^3)) lim((sqr(2)-sqr(1+cosx))/((sinx)^2)) x->0 怎么计算lim(x->0+)x^(1/2)*ln(sinx)? lim(x趋向0)((sinx/x)+1)=2? lim(x-->0) (e^(2x)-1)/sinx Lim x趋向零 (sinx/x)^1/2 求极限lim(x→0) (1-sinx)^(2/x) 求极限 lim(x→0)(1-sinx)^(2/x) x趋于0 lim sinx-tanx/x^3=lim sinx/x×(cosx-1)/(x^2×cosx)是怎么通分的?x趋于0 lim sinx-tanx/x^3=lim sinx/x×(cosx-1)/(x^2×cosx)thank you 1.lim sinX^3/(sinX)^3x→02.lim tan2X/sin5Xx→03.lim(1/1+x)^(1/2x+1)x→0