lim (2cosx)/(3x-π) x接近π/3 求极限 详解过程 谢谢

lim(x_sin)/(3x+2cosx) 求极限lim(x+3)(2+cosx)/(x^2+x) lim(x->0)[cosx^(1/2)]^(π/x) x趋于0 lim sinx-tanx/x^3=lim sinx/x×(cosx-1)/(x^2×cosx)是怎么通分的?x趋于0 lim sinx-tanx/x^3=lim sinx/x×(cosx-1)/(x^2×cosx)thank you lim (2cosx)/(3x-π) x接近π/3 求极限 详解过程 谢谢 lim(x→π/3)sin(x-π/3)/(1-2cosx) lim(x→π/3)sin(x-π/3)/(1-2cosx)= 初学者 求极限lim(x→3)cosx 为什么lim sinx/(x-π) =lim cosx/1 求lim(1-cosx)/x^2求lim(1-COSX)/X^2X→0 求极限lim(x->0) 1/x^3 {[(2+cosx)/3]^x-1}求极限lim(x->0)1/x^3 {[(2+cosx)/3]^x-1} 用洛必达法则计算极限：limx→0 (e^x-cosx)/(sinx)lim x→π (sin3x)/(tan5x);lim x→0 (e^x+e^[-x]-2)/(1-cosx) ;lim x→0 [x(e^x+1)-2(e^x-1)]/x^3 ;lim x→π/2 (Insinx)/(π-2x)^2 lim x→-2 (tanπx)/(x+2) 2.lim x→0 cot2xcot[(π/2)-x] 3.lim x→π/3 (1-2cosx)/(π-3x) 极限lim(2x^2+x)(sinx+cosx)/x^3-2x+1= lim(x->∞)[ xsin(2/x)+(cosx^2)/(x^2)] lim(x→0)(cosx-cos^2x)/x^2 请问高手 x趋近于0 lim (sinx-x*cosx)/(sinx)^3 能不能这样计算原式=lim sinx/(sinx)^3-lim (x*cosx)/(sinx)^3=lim 1/(sinx)^2-lim cosx/(sinx)^2=lim (1-cosx)/(sinx)^2=lim (x^2/2)/(x^2)=1/2,如果可以为何与利用麦克劳林公式所得 求函数的极限 lim cosx/（cosx/2-sinx/2） x->π/2