Question

lim(x趋向正无穷)根号[(x+p)(x+q)]-x怎么做?

Answer 1

分子有理化，得lim(x→＋∞) √[(x+p)(x+q)]-x＝lim(x→+∞) (x+p)(x+q)-x²/{√[(x+p)(x+q)]+x}
=lim(x→+∞) (p+q)/{√[(1+p/x)(1+q/x)]+1}=lim(x→+∞) (p+q)/[√(1*1)+1]=(p+q)/2

Answer 2

=px+qx+pq
∴原式=lim(x→∞)(px+qx+pq)/[√(x+p)(x+q)+x]
因为x→∞
所以只要看x前面的系数的
∴
=（p+q)/上下乘√(x+p)(x+q)+x
则分母就是√(x+p)(x+q)+x
分子打开=x²+px+qx+pq-x²

Answer 3

lim(x→∞) √[(x+p)(x+q)]-x （分子有理化）
=lim(x→∞) {√[(x+p)(x+q)]-x }{√[(x+p)(x+q)]+x }/{√[(x+p)(x+q)]+x }
=lim(x→∞) [(p+q)x+pq]/{√[(x+p)(x+q)]+x }
=(p+q)/2