X->0 (sinx)^2/(1-cosx+sinx) 的极限,

来源：学生作业帮助网 编辑：作业帮 时间：2024/04/25 12:11:55
X->0 (sinx)^2/(1-cosx+sinx) 的极限,

X->0 (sinx)^2/(1-cosx+sinx) 的极限,
X->0 (sinx)^2/(1-cosx+sinx) 的极限,

X->0 (sinx)^2/(1-cosx+sinx) 的极限,
方法一：0/0型极限,用L'Hospital法则
lim(x→0) sin²x/(1-cosx+sinx)
=lim(x→0) (sin²x)'/(1-cosx+sinx)'
=lim(x→0) (2sinxcosx)/(sinx-cosx)
=0
方法二：
lim(x→0) sin²x/(1-cosx+sinx)
=lim(x→0) [2sin(x/2)cos(x/2)]²/[2sin²(x/2)+2sin(x/2)cos(x/2)]
=lim(x→0) [2sin(x/2)cos²(x/2)]/[sin(x/2)+cos(x/2)]
=0

∫(1+sinx) / cos^2 x dx 3sinx-2cosx=0 (1)(cosx-sinx)/(cosx+sinx)+(cosx+sinx)/(cosx-sinx) (2)sin^2 x-2sinxcosx+4cos^2 x cos^2x sinx 不定积分 cos^2x sinx 不定积分 y=(2sinx-1)/(sinx+3)和y=cos^2x+sinx的值域 cos (x/2) - sinx=0求x cos 2x=1-2 sinx 要解x lim（cos(1/x)+2/sinx-1/ln(1+x)) x趋近于0 y=cos^2x+sinx+1(|x| 求函数f(x)=cos^2x+sinx-1 sinx*( cos x)^3/(1+sin^2x)的不定积分 sinx+cos∧2x=1求x f(sinx)=1+cos(2x),求f(x), 函数f(x)=2cos^x+sinx-1,x∈[0,π/2]的值域 化简sin^4x/sinx-cosx - (sinx+cosx)cos^2x/tan^2x-1 f(sinx+1/sinx)=csc^2x-cos^2x 求f(x) 若 f(sinx+1/sinx)=csc^2x-cos^2x,求f(x) 若sinx+sin^2x=1,则cos^2x+cos^4x=?由 sinx+sin^2x=1得 sinx=1-sin^2x=cos^2xcos^2x+cos^4x=cos^2x(1+cos^2x)=sinx(1+sinx)=sinx+sin^2x=1cos^2x(1+cos^2x)=sinx(1+sinx)怎么来的?