求lim(x→0+) ( 2/π*cosπ/2(1-x))/x的极限
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/11 19:46:43
求lim(x→0+) ( 2/π*cosπ/2(1-x))/x的极限
求lim(x→0+) ( 2/π*cosπ/2(1-x))/x的极限
求lim(x→0+) ( 2/π*cosπ/2(1-x))/x的极限
lim(x→0+) { 2/π*cos[(π/2)(1-x)] } / x
= lim(x→0+) cos (π/2 - πx/2) / (πx/2)
= lim(x→0+) sin (πx/2) / (πx/2)
x→0+ 时 πx/2→0
则 lim(x→0+) sin (πx/2) / (πx/2)=1
0/0型,直接罗比达法则一次即可
lim(x→0+) ( 2/π*cosπ/2(1-x))/x
=lim(x→0+) sinπ/2(1-x)/1
=1
求lim(x→0+) ( 2/π*cosπ/2(1-x))/x的极限
求极限 lim(x→0)(lim(n_∞)cos(x/2)cos(x/2²)...cos求极限 lim(x→0)(lim(n_∞)cos(x/2)cos(x/2²)...cos(x/2∧n))
求lim(x→0)x cos 1/x lim(x→∞)x^2/ (3x-1)的极限
求lim(x→0) (cosx+cos^2+.+cos^n-n)/cosx-1 不能用洛必达法则求lim(x→0) (cosx+cos^2+.+cos^n-n)/cosx-1 不能用洛必达法则
如何求 lim(n→∞)cos(x/2)cos(x/4)...cos(x/2n)
求极限lim(x→0) cos(1/x)
求极限:lim(x→0)ln(1+x²)/ (sec x- cos x)
求极限lim(x→0)(1-根号cosx)/[x(1-cos根号x)]
求lim(x→0) (1-cos x) /x^2的极限
求极限:lim(x→1/2) (2x-1)cos(1/(2x-1))
求极限lim(x趋向于0+)(cos根号x)^π /x
x趋近于+0求极限lim(cos√x)^π/x
求lim x趋近于0+ (cos根号x)^(π/x)的极限值..
lim x 趋向于0 时,求 cos(x^2)/x 极限
lim(x→0)(cosx-cos^2x)/x^2
lim( cos(a+2x)-2cos(a+x)+cosa)/x^2x->0 求极限
求极限x→0 lim (1-cos ax)/sin^2x (a为常数)需要过程
lim(x→0) (1-cos(1-cosx))/(arctanx^2)^2求极限