Question

极限计算(不用罗毕达法则)(0004)

Answer 1

∵lim(x->0)[lncos(mx)/x²]

=ln{lim(x->0)[cos(mx)]^(x²)}

=ln{lim(x->0)【{(1+cos(mx)-1)^[1/(cos(mx)-1)]}^[(cos(mx)-1)/(x²)]】}

=lne^{lim(x->0)[(cos(mx)-1)/(x²)]} (利用极限lim(x->0)[(1+x)^(1/x)]=e)

=lim(x->0)[(cos(mx)-1)/(x²)]

=lim(x->0){(-m²/2)[sin(mx/2)/(mx/2)]²}

=-m²/2 (应用极限lim(x->0)(sinx/x)=1)

同理可得lim(x->0)[lncos(nx)/x²]=-n²/2

∴原式=lim(x->0){[lncos(mx)/x²]/[lncos(nx)/x²]}

=lim(x->0)[lncos(mx)/x²]/lim(x->0){[lncos(nx)/x²]

=(-m²/2)/(-n²/2)

=m²/n².