lim x→1 (1-x)tan[(πx)/2]用洛必达法则如何求解我是这样解的：原式=(1-x)/1/tan(πx/2) =-1/1/(π/2)*[sec(πx/2)]^2 =-(π/2)/[cos(πx/2)]^2 =无穷大?答案是2/π 请问正确做法是什么?

lim(x→1) (1-x)tan(π/2)x x→1,求lim（tanπx/4）^tanπx/2求极限, lim(x→0)tan(x)ln(1+x)=? lim(2-x)^tanπx/2 x趋近1 1.计算lim(1/x2-cot2x) X→0 2 lim(1-x)tanπx/2x→1 (1-x)tan(πx/2),lim 趋向于1, lim(x→1)(x-1)tanπx/2=?请不要用洛必达法则. lim（x→1）（1-x^2）tanπx/2 求lim(1-x)tanπx/2 x→1 求x→1时lim（2-x）^tan（πx）/2的极限 lim(1+tan×)^1/x求极限 lim[x→1] tan(x-1)-sin(x-1) / x-1= lim(x→0) [1-cos(x^2)]/sin(x^2)tan(x^2)]=? 求极限：lim(1-x)tan Π/2 几道极限题!1,lim(x->1)(1-x)tanπ/22,lim(x->0)(tanx-sinx)/x^3 lim(2-x)tanπ/4x lim (x→0) [tan( π/4 - x )]^(cotx)=? lim x→-2 (tanπx)/(x+2) 2.lim x→0 cot2xcot[(π/2)-x] 3.lim x→π/3 (1-2cosx)/(π-3x)