Question

求极限:lim(x→0)(tanx-sinx)/x^3

Answer 1

那我就不用洛必达法则了呵呵~,用定理lim[x→0] sinx/x=1
lim[x→0] (tanx-sinx)/x³
=lim[x→消空0] (sinx/cosx-sinx)/x³
=lim[x→0] (sinx-sinxcosx)/(x³cosx)
=lim[x→0] sinx(1-cosx)/(x³cosx)
=lim[x→0] sin³x(1-cosx)/(x³sin²xcosx)
=lim[x→0] (sinx/x)³·(1-cosx)/(sin²xcosx)
=lim[x→0] (sinx/x)³·(1-cosx)/[(1-cos²x)cosx]
=lim[x→0] (sinx/x)³·(1-cosx)/渣桥州[(1+cosx)(1-cosx)cosx]
=lim[x→0] (sinx/x)³如蔽·1/[(1+cosx)cosx]
=1·1/(1+1)
=1/2

Answer 2

【罗必塔漏培法则首芦】
lim(x→0)(tanx-sinx)/x^3
=lim(x→0)(sec^2 x - cosx)/3x^2
=lim(x→者搜带0)(2sec^2 xtanx + sinx)/6x
=lim(x→0)(2sec^2 x/cosx + 1)*sinx/6x
= 3*(1/6)
= 1/2