lim x->0 cos(2x)-1 / cosx -1 =

lim x趋向0 (1+cos^2(1/x))^x lim x->0 cos(2x)-1 / cosx -1 = lim[cos(1/x)]^2,x--->0,极限为什么不存在 求lim(x→0)x cos 1/x lim(x→∞)x^2/ (3x-1)的极限 lim x->0 1-cos(3x)/(2x) 和 lim x->0 1-cos3x/2x^2 lim(x→0)(1/x)*cos(1/x) lim(x→0)(cosx-cos^2x)/x^2 求函数极限：lim(x->0) (cos x)^(1/x) lim((cos( √x))^(1/x)) x……0+ lim(sin(x^2*cos(1/x)))/x怎么做? 求lim (sec^2 x求极限lim (sec^2 x - 1 )/(1-cos x) X-0请写出过程, 用罗必达法则解 （见图）lim [(e^x)-1/sin（5x)]x->0lim [1-cos(5x)/1-cos(7x)]x->0lim x->0 [(8^x-7^x-1)/(x^2)-1]lim [1-(6/x)]^x x->∞ lim 7cos(3x)sec(9x)x->pi/2 极限题,1.lim x-->0 cos(2x)-cos(3x)/x^22.lim x-->∞ cos(2x)-cos(3x)/x^2 lim(x→0) [1-cos(x^2)]/sin(x^2)tan(x^2)]=? lim(x趋近于0)(（1/sin^2 x）-(cos^2 x)/x^2)= (x^2cos^2x-sin^2x)/x^2cos^2xsin^2x （x->0）lim (x^2cos^2x-sin^2x)/x^2cos^2xsin^2x （x->0）=lim [cos^2x-(sin^2x/x^2)]/cos^2xsin^2x=lim (cos^2x-1)/sin^2x=lim -sin^2x/sin^2x=-1这样解错在哪啊?小弟感激不尽 利用lim sinx/x =1或等价无穷小量求极限 lim (cosαx-cosβx)/x^2 x趋向0 求极限lim(x→0) cos(1/x)