lim[x→0] (sin x / x)的 [1/x平方] 次方,这个极限怎么求啊?

lim(x→0)×(x-sin²3x)/x lim (x→0+)sin x/x= lim （x→0）x sin 1/x= lim→0 sin^3/x^3 lim x->0 (sin x-tan x)/sin (x^3) lim(x(sin)^2) 求极限 （（sin(x^3+x^2-x)+sin x) /x x→0 已知lim sinx/x=1 lim（x ->0）sin x /x ^3+3x lim (x->0) sin(3x)+4x/xsec(x) lim x趋于0 （x-sin x）/x^3 lim(x→0)[tan(2x+x^3)/sin(x-x^2)] lim (x->0) [tan(tan x)-sin(sin x)]/(tan x -sin x) 证明lim(x→0)x*sin(1/x)=0我的证明如下:lim(x→0)x*lim(x→0)sin(1/x)因为lim(x→0)x为无穷小量.0 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 lim x→∞ (x+sin 2x)/2 lim sin 3x/x-π,(x→π) LIM(X→∞)X*SIN(1/X) 求极限,lim x趋于0 x * sin 1/x