设a≠0,b≠0,求lim(x→0)[ln(cosax)]/[ln(cosbx)]

lim(x→0)[ln(cosax)]/[ln(cosbx)]
这是0/0型极限，应用洛必达法则，求导得
=lim(x→0)[-(sinax)*a/(cosax)]/[-(sinbx)*b/(cosbx)]
=lim(x→0)(tanax)*a/[(tanbx)*b]
运用等价无穷小，tanax~ax,tanbx~bx
=lim(x→0)ax*a/(bx*b)
=a²/b²

lim(x→0)[ln(cosax)]/[ln(cosbx)]
=lim(x→0)[ln(1+cosax-1)]/[ln(1+cosbx-1)]
=lim(x→0)[cosax-1]/[cosbx-1]
(L'Hospital)
=lim(x→0)[cosax-1]'/[cosbx-1]'
=lim(x→0)[a*(-sinax)]/[b*(-sinbx)]
(L'Hospital)
=lim(x→0)[-a^2*cosax]/[-b^2*cosbx]
=a^2/b^2