lim(x->0)[(x^2)*sin(1/x)]/sinx我用等价无穷小来算,因为当(x->0),sinx～x,所以sin(1/x)～1/x,则原式=lim(x->0) (x^2)*(1/x)/sinx=1,对吗︽

lim(x(sin)^2) lim(x趋向0) x^2 / (sin^2) * x/3 lim(x趋向0) x^2 / (sin^2) * x/3 lim(x趋向0)ln(1+sin x)/x^2 求极限lim(x-->0)x^2 sin(1/x), 求极限 lim sin(x^2 * sin (1/x))/x x->0求极限 lim sin(x^2 * sin (1/x))/x x->0 lim x->0 (sin x-tan x)/sin (x^3) lim(x趋于0)(x^2-sin^2xcos^2x)/((x^2)sin^2x) lim x->0(sqrt(sin(1/x^2)) 求极限 lim sin 2x / sin 5x lim arctan x / xx趋近于0 x趋近于0 求极限 （（sin(x^3+x^2-x)+sin x) /x x→0 已知lim sinx/x=1 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->0) 1-x^2-e^(-x^2)/x*sin^(3)2x lim(x→0)[tan(2x+x^3)/sin(x-x^2)] lim（x ->0）sin x /x ^3+3x lim(x→0)×(x-sin²3x)/x lim (x->0) sin(3x)+4x/xsec(x) lim x趋于0 （x-sin x）/x^3