百度作业网求极限lim(x→0)［∫(0.x)sintdt+lncosx］/x∧4
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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
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