Question

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

Answer 1

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