lim t→0 tan(6t)/sin(7t)=?
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/25 22:15:33
limt→0tan(6t)/sin(7t)=?limt→0tan(6t)/sin(7t)=?limt→0tan(6t)/sin(7t)=?t→0则tan(6t)~6t/sin(7t)~7t所以原式=6
lim t→0 tan(6t)/sin(7t)=?
lim t→0 tan(6t)/sin(7t)=?
lim t→0 tan(6t)/sin(7t)=?
t→0 则tan(6t)~6t
/sin(7t)~7t
所以原式=6/7
lim t→0 tan(6t)/sin(7t)=?
lim t-0 求 tan(20t) / sin(4t)
Lim t-->0 (tan 6t)/(sin 2t) 最好有具体步骤和讲解......
t->0,lim[tan(sinx)-sin(tanx)]/(tanx-sinx)=?
lim[t→0+](−2sin(6t)/sin(6t)+2tcos(6t))百思不得其解,求大神出手相助……
解一道大一极限题 lim(x→1)(1-x^2)/sinπxt=1-x,t-->0lim(2t-t^2)/sin(π-πt)=lim(2-t)t/sinπt=lim(2-t)t/πt=2/πlim(2-t)t/sinπt=lim(2-t)t/πt=2/π这部怎么来的
这道数学题是怎么变化出来的?lim sin(3兀+3t)=?x->0 lim tan(5兀+5t)=?x->0
lim △t→0
lim(x->0)∫sin(sqrt(t))dt/x^a,
最后一道题了,计算lim【x→0】 (∫【0→x^2】sin(t^2)dt)/x^6)
lim x→0∫sin(t^2)dt]/(x^6-x^7)上限为x^2,下限为0
x=tan(t) sin(t)-cos(t)=?
lim (x->0) [tan(tan x)-sin(sin x)]/(tan x -sin x)
lim x->0 (sin x-tan x)/sin (x^3)
设cosθ=t(t≥0),求sinθ和tanθ的值
求极限,lim(t→0)[(t+a)^t-1]/t
高数,求极限的问题 lim [tan(tanx)-sin(sinx)] / (tanx高数,求极限的问题lim [tan(tanx)-sin(sinx)] / (tanx - sinx)x→0
t趋近于0时,求(1 1/t)t的极限(注:t表示t次方)lim(1+1/t)^t(t→0)=e是怎么得来的呢?我记得应该是lim(1+1/t)^t(t→∞)=e (等同于lim(1+t)^(1/t)(t→0)=e)难道lim(1+1/t)^t(t→0)与lim(1+1/t)^t(t→∞)相等?敬请