lim(x->PI/2) COS X/(PI/2)-X
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/16 05:24:32
lim(x->PI/2) COS X/(PI/2)-X
lim(x->PI/2) COS X/(PI/2)-X
lim(x->PI/2) COS X/(PI/2)-X
lim(x->PI/2) COS X/(PI/2)-X =lim(x->PI/2) sin(PI/2-X/(PI/2)-X =lim(x->0) sinx/x=1
你好!
用罗比达法则
原式=lim
= 1
lim(x->PI/2) COS X/(PI/2)-X
lim(x->0+)(cos√x)^pi/x
Lim[(3x)^1/2] * e^(cos(8pi/ x) ( x-->+0) 求极限?
用罗必达法则解 (见图)lim [(e^x)-1/sin(5x)]x->0lim [1-cos(5x)/1-cos(7x)]x->0lim x->0 [(8^x-7^x-1)/(x^2)-1]lim [1-(6/x)]^x x->∞ lim 7cos(3x)sec(9x)x->pi/2
y=cos(-pi/2x+pi/2)等于什么
求lim x->0 cos(PI/x) / x该题的原意是,当x→0时,证明cos(PI/x) / x不是无穷大量
化简sin(2x-pi)sin[x+(3/2)pi]/cos[x+(5/2)pi][1-sin(pi+x)]
证明lim(x->负无穷)arctanx=-pi/2
x→pi/2 lim tanx/tan3x=?
函数f(x)=sin^2(x+pi/12)+cos^2(x-pi/12)的最大值
f(2x)=sin (x+3*pi/4)+cos (x-pi/2),
2(sin^4 x + cos^4 x) =1 [ - pi ≤ x ≤ pi ]
Matlab u(x,t)=sin(5*pi*x)cos(5*pi*t)+2sin(7*pi*x)cos(7*pi*t) 其中0
f(x)h(x)=√2/2sin(x+pi/4)cos(x+5pi/4) =-√2/2sin(x+pi/4)cos(x+pi/4)√2/2sin(x+pi/4)cos(x+5pi/4) =-√2/2sin(x+pi/4)cos(x+pi/4)这个是怎么化出来的?
sin(pi/2+x)=?如题哈:(1)sin(pi/2+x)=cos(x) ------> (2)cos(pi/2+x)=-sin(x) ---------->
sin(2x+pi/3)=cos(2x-pi/6) 成立吗
y=sin(2x+Pi/3)+cos(Pi/6-2x)周期为
函数y=sin(pi/2+x)cos(pi/6+x)的最大值