设a>0,lim（x→＋∞）[ax-sql(x^2+bx)]=1,求常数a、b的值

1 = lim（x→＋∞）[ax-sql(x^2+bx)]
= lim（x→＋∞）{[(ax)^2 - (x^2+bx)]/[ax + sql(x^2+bx)]}
= lim（x→＋∞）{[(a^2 - 1)x^2 - bx)]/[ax + sql(x^2+bx)]}
= lim（x→＋∞）{[(a^2 - 1)x - b)]/[a + sql(1+b/x)]}
因a>0,分母的极限为 a + 1.要使整个分式有极限,只有,
a^2 - 1 = 0,a^2 = 1,a = 1.
此时,
1 = lim（x→＋∞）{[ - b]/[1 + sql(1+b/x)]}
= [ - b]/[1 + sql(1+0)]
= -b/2
b = -2.
因此,
a = 1,b = -2.