lim(x—>∞)2xsinx/secx—1 说下步骤 谢谢

lim(x->0)(2xsinx)/(secx-1) lim(x—>∞)2xsinx/secx—1 说下步骤 谢谢 用洛必达法则求lim x→0 tanx-x /(x²sinx)的极限lim x→0 secx/x²-1/xsinx (无穷小代换）=lim x→0 1/x²(secx-1） （洛必达法则）=lim x→0 secxtanx/2x=lim x→0 cosx/2=1/2lim x→0 secx/x²-1/xsinx (无穷小代 lim x→∞ [(secx)^2-1]/3x^2=2tanx(secx)^2/6xlim x→∞ [(secx)^2-1]/3x^2如何能化成lim x→∞ 2tanx(secx)^2/6x lim(x→-∞）e^xsinx lim(x趋于0时)secx - 1/x^2 , lim(x→0) xsinx lim(1+cosx)的2secx次方 当x—>π/2 （派.的极限是多少~ 极限lim(x趋进0)(1-cosx)^2secx lim(x→0)[(secx)^2-1] /(1-cosx) lim（x→0）（secx）^2tanx lim x→0(1+2cosx)^secx lim (1/(xsinx)-1/x^2)x~0 lim(e^2-x)/xsinx x→0 怎么求? 极限lim x->0 (xsin2/x+2/xsinx)=? lim(xsinx)/x^2-4,x趋向于无穷 lim xsinx/(x^2-4) x趋向正无穷要用迫敛性证, 求lim(x→∞)1-cos2x/xsinx