lim(x→0+)[ln(sin5x)]/[ln(sin3x)] =lim(x→0+)[5cos5x/sin5x]/[3cos3x/sin3x] =lim(x→0+)[5cos5x×sin3x]/[3cos3x×sin5x]=1（L'Hospital法则）补充：lim(x→0+)cos5x=1,lim(x→0+)cos3x=1,因为这是连续函数求极...