limx→π/2(ln(sinx))/(π-2x)^2求极限
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/07/08 13:47:07
limx→π/2(ln(sinx))/(π-2x)^2求极限
![limx→π/2(ln(sinx))/(π-2x)^2求极限](/uploads/image/z/5867731-19-1.jpg?t=limx%E2%86%92%CF%80%2F2%28ln%28sinx%29%29%2F%28%CF%80-2x%29%5E2%E6%B1%82%E6%9E%81%E9%99%90)
lim(x→π/2) (ln(sinx))/(π-2x)^2 (0/0)
=lim(x→π/2) cotx/[-4(π-2x)] (0/0)
=lim(x→π/2) -(cscx)^2/8
=-1/8
再问: (π-2x)^2怎么算出-4(π-2x)?
再答: [(π-2x)^2]'
=2(π-2x)(-2)
=-4(π-2x)
再问: cotx=cscx^2?
书上是说1/sinx*cosx=1/sin^2x-csc^2x
再答: (lnsinx)'
= (1/sinx). (sinx)'
= cosx/sinx
=cotx
(cotx)' = -(cscx)^2
再问: cscπ/2=1?
再答: cscx= 1/sinx
csc(π/2) = 1/sin(π/2) = 1
=lim(x→π/2) cotx/[-4(π-2x)] (0/0)
=lim(x→π/2) -(cscx)^2/8
=-1/8
再问: (π-2x)^2怎么算出-4(π-2x)?
再答: [(π-2x)^2]'
=2(π-2x)(-2)
=-4(π-2x)
再问: cotx=cscx^2?
书上是说1/sinx*cosx=1/sin^2x-csc^2x
再答: (lnsinx)'
= (1/sinx). (sinx)'
= cosx/sinx
=cotx
(cotx)' = -(cscx)^2
再问: cscπ/2=1?
再答: cscx= 1/sinx
csc(π/2) = 1/sin(π/2) = 1
limx→π/2(ln(sinx))/(π-2x)^2求极限
limx→π/2 (sinx)^tanx limx→∞(2x+3/2x+1)^x+1 求极限
求极限limx→0(x-sinx)/x^2
limx→0 (1/x^2)ln(sinx/x)求极限,ln和前面的(1/x^2)是相乘的关系~
limx趋于0,ln(1-2x)/sinx,求极值
limx ln(1+1/x^2) 求极限?
求极限limx趋于0(ln(1+2x))/tan5x
limx->π[sinx/(π-x)]求极限
极限limx→0 x/ln(1+x^2)=()
求极限limx->0(sinx/x)^(1/x^2)
求极限lim(x→π/2) ln(sinx)/[(π-2x)^2]怎么计算?
limx~0,x/(sinx∧2)的极限 怎么求?