求lim(x→π/4) (sin2x-cos2x-1)/(cosx-sinx)的极限
求lim(x→π/4) (sin2x-cos2x-1)/(cosx-sinx)的极限
lim cos2x/(sinx-cosx) x→π/4 求函数的极限
极限的运算法则 求(sin2x-cos2x-1)/(cosx-sinx)在x趋近于π/4时的极限
已知x∈(0,π),且sinx+cosx=1/2 求sin2x+cos2x,sinx-cosx的值
sinx+cosx=1/5,0<x<π,求sinx、cosx、tanx,求sin2x-cos2x的值
lim(x→π/4)(sinx-cosx)/cos2x
sinx+cosx=1/5,x属于(π/2,3π/4),求sinxcosx,sin2x,cos2x,sinx,cosx
求极限 lim(cosx+sinx)^1/x
求解一道极限题lim【(cos2x)的1/sin2x次方】=?(x→0)
急.求极限 lim x趋向于π/2,(sinx)的(1/cosx)^2
lim(√1+cosx)/sinx x趋于π+ 求极限
求函数的极限 lim cosx/(cosx/2-sinx/2) x->π/2