lim(x趋向正无穷)根号[(x+p)(x+q)]-x怎么做?
问题描述：
lim(x趋向正无穷)根号[(x+p)(x+q)]-x怎么做?
答
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
答
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