求极限lim(x→0)［∫(0.x)sintdt+lncosx］/x∧4
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/19 13:15:56
求极限lim(x→0)［∫(0.x)sintdt+lncosx］/x∧4求极限lim(x→0)［∫(0.x)sintdt+lncosx］/x∧4求极限lim(x→0)［∫(0.x)sintdt+lnc
求极限lim(x→0)［∫(0.x)sintdt+lncosx］/x∧4
求极限lim(x→0)［∫(0.x)sintdt+lncosx］/x∧4
求极限lim(x→0)［∫(0.x)sintdt+lncosx］/x∧4
lim(x→0)［∫(0.x)sintdt+lncosx］/x∧4
=lim(x→0)［-cosx+1+lncosx］/x∧4
=lim(x→0)［sinx-sinx/cosx]/(4x^3)
=lim(x→0)［cosx-1/cos^2x]/(12x^2)
=lim(x→0)［-sinx-2sinx/cos^3x]/(24x)
=lim(x→0)［-cosx-(2cos^3x-6sin^2x)/cos^5x]/24
=[-1-(2-0)/1]/24
=-1/8
=lim(sinx-sinx/cosx)/4x^3 =lim(sinx/x)(cosx-1)/(cosx*4x^2) =lim(cosx-1)/4x^2) =lim-sinx/8x=-1/8
求极限lim(x→0)［∫(0.x)sintdt+lncosx］/x∧4
求极限lim Ln(1+x) /x > .< x→0
求极限lim(x→0)x-sinx/x^3
lim x-0 sin2x/x 求极限
lim(x->0)arctan1/x求极限
求极限lim(x->0)x^sinx,
求极限lim(x→0)sinxsin(1/x);lim(x→∞)(arctanx/x)
求极限 lim(sin2x/sin3x)x->0x→0
求极限:lim(x→0)(tanx-sinx)/x^3
求极限lim(x→0)(tanx-sinx)/(x-sinx)
求极限lim(x→0)x²/1-cosx
求极限：lim x→0 xsin1/x²
求lim(x→0)arctanx/x的极限,
求极限lim x→0 (lnx)/x
求极限lim(x→0) cos(1/x)
求极限lim(x→0)sin3x/x
lim(x→0)tan3x/x的极限怎么求?
求极限lim(x→0)[tan3x/x]