求极限x→0 lim (1-cos ax)/sin^2x (a为常数)需要过程
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/11 20:10:31
求极限x→0 lim (1-cos ax)/sin^2x (a为常数)需要过程
求极限x→0 lim (1-cos ax)/sin^2x (a为常数)需要过程
求极限x→0 lim (1-cos ax)/sin^2x (a为常数)需要过程
x→0 lim (1-cos ax)/(sinx)^2
=x→0 lim (a*sin ax)/(2sinx*cosx)
=x→0 lim (a*sin ax)/sin2x
=x→0 lim (a^2*cos ax)/2cos2x
=(a^2*1)/(2*1)
=(a^2)/2
就是用两次洛必达法则就行了
∵将0分别代入分子分母均为0可知为0/0型不定式
∴由洛比达法则得x→0lim【(1-cos ax)/sin^2x 】=x→0lim[(asinax)/(2cos2x)]=0
=lim (1/2)(ax)^2 / x^2 【等价无穷小代换:u→0时,1-cos u ~ (1/2)u^2; sin u ~ u】
= a^2 / 2
求极限lim(x→0) cos(1/x)
求极限x→0 lim (1-cos ax)/sin^2x (a为常数)需要过程
求导数,求极限Lim x取近于0 ln(cos ax)/ln(cos x)
求函数极限:lim(x->0) (cos x)^(1/x)
求lim(x→0)x cos 1/x lim(x→∞)x^2/ (3x-1)的极限
求函数极限lim x→0 e^x-x-1/x cos x
求极限:lim(x→0)ln(1+x²)/ (sec x- cos x)
求极限lim(x->0)(1+ax)^(1/bx)
求极限lim(x→0)(1-根号cosx)/[x(1-cos根号x)]
求lim(x→0+) ( 2/π*cosπ/2(1-x))/x的极限
求lim(x→0) (1-cos x) /x^2的极限
求极限lim(x→∞) ((x²+1)/(x-1)-ax-b)=0
求极限lim(1-√cosx)/(1-cos√x) (x→0+)
求极限lim(x,y)→(0,0) [1-cos(x2+y2)]/[(x2+y2)e^x2y2]
lim(x→0) (1-cos(1-cosx))/(arctanx^2)^2求极限
求lim (sec^2 x求极限lim (sec^2 x - 1 )/(1-cos x) X-0请写出过程,
高数极限高数极限高数极限请用洛必达法则求极限lim x→0 ln(1-ax)/ln(1+bx)=?
求极限lim(x→∞) ((x²+1)/(x-1)-ax-b)=0求a,b求极限lim(x→∞) ((x²+1)/(x-1)-ax-b)=0求a,b