1) lim (x→3pi/2) [cos7x/(x-3pi/2)] 答案不是+∞或-∞
来源:学生作业帮 编辑:大师作文网作业帮 分类:综合作业 时间:2024/11/11 17:28:46
1) lim (x→3pi/2) [cos7x/(x-3pi/2)] 答案不是+∞或-∞
2) lim→0 [sin^2 (5teta) / (5teta) ]
有图
2) lim→0 [sin^2 (5teta) / (5teta) ]
有图
用罗必达法则:
1.lim (x→3pi/2) [cos7x/(x-3pi/2)]
= lim (x→3pi/2)(-sin7x * 7)/1 = -7sin(21pi/2) = -7
2.lim→0 [sin^2 (5teta) / (5teta) ]
= lim→0 (2sin(5theta) * cos(5theta)* 5/5 = 0
1.lim (x→3pi/2) [cos7x/(x-3pi/2)]
= lim (x→3pi/2)(-sin7x * 7)/1 = -7sin(21pi/2) = -7
2.lim→0 [sin^2 (5teta) / (5teta) ]
= lim→0 (2sin(5theta) * cos(5theta)* 5/5 = 0
lim x->pi (x^2-1)/cosx
Lim[(3x)^1/2] * e^(cos(8pi/ x) ( x-->+0) 求极限?
已知函数f(x)=cos(2x-pi/3)+2sin(x-pi)*sin(x+pi/4)
若sin((pi/6)+x)=1/3,则cos((pi/3-x)=?cos((2pi/3)+x)=?
化简sin(x+PI/3)+2sin(x-PI/3)-根号3*cos(2pi/3-1)
已知Sin(2x+pi/3)=1/3,求Sin(5Pi/6-4x),
lim (sec x - tan x) limit是x->(pi /2 )-
lim(x→pi)x^cosx
已知函数f(x)=sin(wx+pi/3),w>0,且f(pi/6)=f(pi/2),函数在(pi/6,pi/2)上有最
定积分,从-pi/2到pi/2,∫x*(cosx)^(3/2) dx=?
已知向量a=(sinx,根号3),b=(1,cosx),x属于(-pi/2,pi/2).
证明lim(x->负无穷)arctanx=-pi/2