ln(1-sin^2x)/(1+cosx)(e^x^2-1)当x→0的极限

cosx ^ (1/ln(1+x^2))怎么化为e^(ln cosx/ln(1+x^2)) lim{[sinx+(x^2)*sin(1/x)]/[(1+cosx)ln(1+x)]}(x趋近于0） lim((sinx-ln(1+x^2)^1/2)/x^4) x->0求解 呵呵~sin改成-cosx+1 ln(1-sin^2x)/(1+cosx)(e^x^2-1)当x→0的极限 求极限 x趋向于π/3 ((1-2cosx)ln(1+x))/sin(x-π/3) ∫secx dx=?∫secx dx=∫(dx)/cosx=∫(cosx/cos²x)dx =∫(d sinx)/(1-sin²x) =(1/2)ln│(1+sinx)/(1-sinx)│+C =(1/2)ln(1+sinx)²/(1-sin²x)+C =(1/2)ln[(1+sinx)/cosx]²+C =ln│secx+tanx 2ln|sec X+tan X|=ln|(1+sin x)/(sin x-1)|怎么证 求 lim ln(1+x+2x^2)+ln(1-x+x^2)/secx-cosx lim(x→0)(cosx)^(1/ln(1+x^2)) 化简(sin^2 x/sin x-cosx)-(sin x+cosx/tan^2 x-1) (sin^x/sinx-cosx)-sinx+cosx/tan^2x-1 lim (cosx)^ln(1/1+x^2)怎么算 lim(sin²x-x²cosx))/(x²ln(1+x)arcsinx) 当x趋近于0时 limx趋向0[ln(1+x^2)/secx-cosx] ∫cosX*ln(1+X^2)dx怎么求? 求∫cosX*ln(1+X∧2)dx? lim(x→0) (ln cosx)/[ln(1+x^2)] 等于多少? 一道利用等价无穷小求极限的题lim[sinx+(x^2)sin(1/x)]/[(1+cosx)ln(1+x)] x趋近于0