lim[tan(π/3+h)-tan(π/3-h)]/h h→0求极限

lim[tan(π/3+h)-tan(π/3-h)]/h h→0求极限 lim(2-x)tanπ/4x 求极限 lim(n→∞) tan^n (π/4 + 2/n) lim(n→∞)tan^n(π/4+2/n) =lim(n→∞)[(tan(π/4)+tan(2/n))/(1-tan(π/4)tan(2/n))]^n =lim(n→∞)[(1+tan(2/n))/(1-tan(2/n))]^n =lim(n→∞)(1+tan(2/n))^n/(1-tan(2/n))^n (1) 因为 lim(n→∞)(1+tan(2/n) tan=3,π tan(-20/3π) x→1,求lim（tanπx/4）^tanπx/2求极限, 求极限：lim(1-x)tan Π/2 lim (x→0) [tan( π/4 - x )]^(cotx)=? (1-x)tan(πx/2),lim 趋向于1, lim(x→1) (1-x)tan(π/2)x lim(2-x)^tanπx/2 x趋近1 几道极限题!1,lim(x->1)(1-x)tanπ/22,lim(x->0)(tanx-sinx)/x^3 求lim(x->1)[(3x-3)tan((π/2)x] 已知β-a=γ-β=π/3,求tanαtanβ+tanβtanγ+tanγtanα的值 tan(3π/5)和tan(-3π/5)大小 比较tan(-π/5)与tan(-3π)的大小 tan(α-3π/2)乘tan(2π-α) tan(4/π+α)=3 求tanα?