x*2-3x+2/sin(x-1)在x=1处的极限

傅里叶级数作图f（x）=2sin[x] - sin[2x] + 2/3sin[3x] - 1/2sin[4x]我用mathematica输入程序Plot[{2sin[x],-2sin[x],2sin[x] - sin[2x],-2sin[x] + sin[2x],2sin[x] - sin[2x] + 2/3sin[3x],-2sin[x] + sin[2x] - 2/3sin[3x],2sin[x] - sin[2x] + 2/3si 求极限 （（sin(x^3+x^2-x)+sin x) /x x→0 已知lim sinx/x=1 x*2-3x+2/sin(x-1)在x=1处的极限 数学达人帮助求极限!当x->0时sin(x)=x+o(x)x*cos(x)=x-x^3/2!+o(x^3)sin(x)^3=x^3+o(x)lim(sin(x)-x*cos(x)/sin(x)^3=1/2请问错在哪里?谢谢! 2sin^2x - 5sin x cos x=3 cos^2x,求X, 已知sin(x)+3cos(x)=2,求sin(x)-cos(x)/sin(x)+cos(x)=? (1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x 化简[1-(sin^4 x-sin^2 xcos^2 x+cos^4 x)]/(sin^2 x)+3sin^2 x s = 2*sin(x)-sin(2*x)+2/3*sin(3*x)-1/2*sin(4*x)+2/5*sin(5*x)用matlab画图,求教啊 x^ 3(sin x )^2/x^ 4+2x +1在[-1,1]的定积分 求证(3-sin^4 x-cos^4 x)/2cos^2 x=1+tan^2 x+sin^2 x f(x)={(1-e^x)/sin(x/2) ,x>0 ;ae^2x,x 三角等式求证：cos^6x+sin^6x=1-3sin^2x+3sin^4x (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 已知tan=2,求(cos x+sin x)/(cos x-sin x)+sin^2x 化简(sin x+cos x-1)(sin x-cos x+1)/sin 2x 求证（cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2x 求导f(x) = 4x^3 + sin(x^2)