lim(2/π·arcosx)^1/x,x趋近0,洛必达法求?

lim(2/π·arcosx)^1/x,x趋近0,洛必达法求? 洛必达求极限 limsinxlnx x趋近于0+,lim(2/π·arctanx)^x x趋近无穷大,lim(ln1(/x))^x x趋近0+,limlnx·ln(1+x) x趋近0+,lim(x^3+x^+x+1)^1/3-x x趋近无穷大,lim (e^x-e^sinx)/(x-sinx) x趋近0,lim(sinx/x)^(1/x^2)x趋近0,lim[1/e· (1/+x) lim(x→1) (1-x)tan(π/2)x lim(2-x)^tanπx/2 x趋近1 lim(1-x)^(2/x) x->0 (1-x)tan(πx/2),lim 趋向于1, lim (π/2-arctanx)/1/x x大于0 lim(x->0)[cosx^(1/2)]^(π/x) 为什么lim sinx/(x-π) =lim cosx/1 lim x(π/2-arctanx)＝?lim下面是+∞ 1.计算lim(1/x2-cot2x) X→0 2 lim(1-x)tanπx/2x→1 求极限 lim x→π/2 (sin2x)^3 lim x→0 根号(1+x^2)-1/x lim(2-x)tanπ/4x 极限：lim(x->2)1/(x-2) lim(1-(5/x))^x-2 lim(x+e^2x)^(1/sinx) 求下列极限：1.lim(x→2)(x+2/x-2) 2.lim(x→0)4x^3-2x^2+x^2-2x)3.lim(x→0)tanx-sinx/x^34.lim(x→π)sin3x/x-π5.lim(x→无穷大量）（1-1/2x)^x+2知道答案,求过程!谢谢 求极限：lim(1-x)tan Π/2