Question

高等数学求极限问题，大一新生跪求解决！

Answer 1

求极限：x→0lim(tanx-sinx)/x³
解一：原式=x→0lim(sinx-sinxcosx)/(x³cosx)=x→0limsinx(1-cosx)/(x³cosx)
=x→0lim(1-cosx)/(x²cosx)=x→0lim[2sin²(x/2)]/[x²(1-2sin²(x/2)]
=x→0lim(x²/2)/[x²(1-x²/2)]=x→0lim[1/(2-x²)]=1/2
解二：原式=x→0lim(sec²x-cosx)/(3x²)=x→0lim(2sec²xtanx+sinx)/(6x)
=x→0lim(4sec²xtan²x+2sec⁴x+cosx)/6=3/6=1/2.

Answer 2

用定理lim[x→0] sinx/x=1
lim[x→0] (tanx-sinx)/x³
=lim[x→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)/[(1+cosx)(1-cosx)cosx]
=lim[x→0] (sinx/x)³·1/[(1+cosx)cosx]
=1·1/(1+1)
=1/2

Answer 3