limx→π/2:(cosx)^(π/2-x)
来源:学生作业帮助网 编辑:作业帮 时间:2024/12/01 12:29:44
x)̭x6|Fr~qf[iTW^Ά DO{~糩T)X
p
a
b!wMy>eӟo}o߳Sv|ټƗ3' D[WvYqFtK*RH2A}Pe
Z
(^EU 2bgkTBx ~qAb(us
limx→π/2:(cosx)^(π/2-x)
limx→π/2:(cosx)^(π/2-x)
limx→π/2:(cosx)^(π/2-x)
y=(cosx)^(π/2-x)
取对数:lny=(π/2-x)lncosx
lny=lncosx/[1/(π/2-x)]
应用罗必达法则求极限:
lny=lim (-sinx/cosx)/[1/(π/2-x)^2]
=lim -tanx (π/2-x)^2
=lim -(π/2-x)^2/cosx
=lim -(π/2-x)/sin(π/2-x) * (π/2-x)
=lim -(π/2-x)
=0
因此y=1
即原式极限=1
limx->π/2 cosx/(x-π/2)
limx→π/2:(cosx)^(π/2-x)
求limx→π/2(2)[sinx-cosx-x^2]
limx→π/2 (2+sinx)/cosx的极限是多少
limx→0 (cosx)^((cosx)^2)
limx-->π/2 (1+cosx)^tanx 求极限
limx→∞(cosx/x^2)
limx→∞(sinx-x^2)/(cosx+x^2)
求极限limx→0(cosx)x^4/2
limx→0 1-cosx/x^2的值
求极限limx→0 (cosx)^1/sin^2x求极限limx→0 (cosx)^(1/sin^2x)
limx趋近于二分之π (1-sinx)/(cosx/2-sinx/2),非常非常感谢
limx趋向0 ln(cosx)/x^2
limx→0(cosx)^(1/x^2)求解lim(x→0)((cosx)^(1/x^2))
求下列极限:1、limx→0 1-cosx/(xtanx) 2、limx→0 [(根号下1+x)-1]/tan2x
求下列极限:1、limx→0 1-cosx/(xtanx) 2、limx→0 [(根号下1+x)-1]/tan2x
limx→1(1-x)^(cosxπ/2)求极限lim(2/π arctanx)^x 其中x趋向于正无穷大
极限,limx→∞cosx