求极限lim(x→0)[∫(0.x)sintdt+lncosx]/x∧4
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/09/21 03:27:23
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(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
求极限lim(x→0)lncosx/xsinx
求上下极限lim(x趋近0){∫(o-x) sintdt}/x
求极限lim(x→0)∫sintdt/x^2上标为x下标为0
求上下极限lim(x趋近无穷大){∫(o到x) sintdt}/x
请问,计算极限lim(x→0) ∫te^tdt变限范围(0,x^2)/∫x^2sintdt变限范围(0,x) 书上第一步
极限x→0,求lim(∫(上x下0)sint^3dt)/x^4
lim(x-0+)x∧(x);用洛必达法则求极限.
求极限lim Ln(1+x) /x > .< x→0
求极限lim x→0 (x-xcosx)/(x-sinx)
求极限lim(x→0)x-sinx/x^3
求极限lim(x->0)x^sinx,
lim x-0 sin2x/x 求极限