用洛必达法则求极限lim(x→0)(sin 4x)/(sin
用洛必达法则求极限用洛必达法则求极限lim(x→∞)(1+sin x)的1/x次幂不好意思,写错了,是X趋向0
求极限lim(x→∞)(sin√x+1-sin√x)
求极限lim(x→1)(1-x^2)/(sinπx)
求极限 lim[(1-x²)/sinπx] x→1
求lim x的sin x次幂的极限(x趋近于0+)
求极限,lim x趋于0 x * sin 1/x
求极限lim(x-->0)x^2 sin(1/x),
求极限lim(x→0)(1/sin²x-1/x²cos²x)
求极限 ((sin(x^3+x^2-x)+sin x) /x x→0 已知lim sinx/x=1
二元函数求极限:lim sin(x^2+y)/(x^2+y^2) x→0,y→0
二元函数求极限:lim (sin(x^2+y)) / (x^2+y^2) x→0,y→0
求极限lim(x→∞)定积分sin^nxdx x∈[0,π/4]