lim (x→0) ［(2x) / (1+x^2)］/sec x tan x+silim (x→0) ［(2x) / (1+x^2)］/sec x tan x+sin x怎么算出lim （x→0）(x/sin x ) （cos x^2/1+cos x^2）（2/1+x^2）

lim(x→0)e^x-x-1/x^2 lim(1-x)^(2/x) x->0 lim(x→0)(1-2x)^1/x 极限? lim 2（1+x）^1/x x→0 极限lim(x→0)(1-2/x)^x lim(x→0) (e^(-1/x^2))/x^100 lim(x→0)(2-e^x)^(1/x) lim(x→0)2x^3-x+1 lim(x→0)(a^2x-1)/4x 求lim(x→0)x cos 1/x lim(x→∞)x^2/ (3x-1)的极限 求lim(1-cosx)/x^2求lim(1-COSX)/X^2X→0 lim(x→0)(1-cos2x)/(x*sin2x) lim(x→0)(xcot2x) lim(x→无穷)(1+(1/2x))^x 1.lim x→0 3x/(sinx-x)2.lim x→0 (1-cosmx)/x^23.lim x→0 sin4x/[(√x+2)-√2]4.lim h →0 [sin(x+h)-sin(x-h)]/h 当x→0时,lim[ln(1+2x)+xf(x)]/x^2=2,求lim[2+f(x)]/x为什么这样解是错的?lim[ln(1+2x)]/x^2+limf(x)/x=lim2x/x^2+limf(x)/x=lim[2+f(x)]/x=2不是有lim[f(x)+g(x)]=limf(x)+limg(x)吗? 当x→0时,lim[ln(1+2x)+xf(x)]/x^2=2,求lim[2+f(x)]/x 为什么这样解是错的?lim[ln(1+2x)]/x^2+limf(x)/x=lim2x/x^2+limf(x)/x=lim[2+f(x)]/x=2不是有lim[f(x)+g(x)]=limf(x)+limg(x)吗? lim(x→0)(cosx)^(1/ln(1+x^2)) lim(x→0)(1+sinx)^(1/2x) lim(x→0)[1/x^2-(cotx)^2]