lim(x→0)(1-cosx)/(xsinx)=?
来源：学生作业帮助网 编辑：六六作业网 时间：2024/11/22 05:02:32
lim(x→0)(1-cosx)/(xsinx)=?lim(x→0)(1-cosx)/(xsinx)=?lim(x→0)(1-cosx)/(xsinx)=?lim(x→0)(1-cosx)/(xsin
lim(x→0)(1-cosx)/(xsinx)=?
lim(x→0)(1-cosx)/(xsinx)=?
lim(x→0)(1-cosx)/(xsinx)=?
lim(x→0)(1-cosx)/(xsinx)
=lim(x→0)(1-(1-2(sin x/2)^2)/(xsinx)
=(1-(1-2*x^2*(1/2)^2))/x^2
=1/2
1-cosx(x->0)等价于x^2/2
sinx/x(x->0)=1
所以1/2
这是0/0型，可以用洛必达法则
lim(x→0)(1-cosx)/(xsinx)
=lim(x→0)sinx/(sinx+xcosx)
=lim(x→0)1/(1+x/tanx)
只需求出lim(x→0)x/tanx即可
这也是0/0型，也可以用洛必达法则
lim(x→0)x/tanx
=lim(x→0)1/(secx)^2
=1
所以原极限=1/2
1/2
求Lim(x→0)(sinx/x)^(cosx/1-cosx)
lim(x→0)(1/cosx)=?
求极限lim(x~0)((e^x+e^2x+e^3x)/3)^1/xs
求lim(1-cosx)/x^2求lim(1-COSX)/X^2X→0
lim（x→0） 1-cosx/x＋x
lim(x→0)(e^x-cosx)/x=1?
lim(x→0)(cosx)^(1/ln(1+x^2))
lim((x→0) (sinx/x)^1/(1-cosx)
求极限lim(x→0)x²/1-cosx
求极限lim(x→0)(cosx-(1/x))
lim(x—0) (1-cosx)/x
limx→0(cosx)^(1/x^2)求解lim(x→0)((cosx)^(1/x^2))
lim(x→0)[(secx)^2-1] /(1-cosx)
lim【x→0+】(1-cosx)^(1/lnx)
数学求极限lim x→0 ( x3/1-cosx)
求下列极限lim(x→0)(1-cosx)/xsinx
lim(x→0)(1-cosx)/(xsinx)=?
求详细解：lim(x→0+) (1-cosx)^sinx