求极限lim(x→0)[∫(0.x)sintdt+lncosx]/x∧4
问题描述:
求极限lim(x→0)[∫(0.x)sintdt+lncosx]/x∧4
答
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