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.