问题描述: lim (x→0)[√﹙x²﹢x+1﹚]-[√﹙x²-x+1﹚]求极限 1个回答 分类:数学 2014-11-19 问题解答: 我来补答 lim (x→0)[√﹙x²﹢x+1﹚]-[√﹙x²-x+1﹚]=0是不是x-->∞[√﹙x²﹢x+1﹚]-[√﹙x²-x+1﹚]={[√﹙x²﹢x+1﹚]-[√﹙x²-x+1﹚]}/1分子分母同时乘以[√﹙x²﹢x+1﹚]+[√﹙x²-x+1﹚]=[(x²+x+1)-(x²-x+1)/[√﹙x²﹢x+1﹚]+[√﹙x²-x+1﹚]=2x/[√﹙x²﹢x+1﹚]-[√﹙x²-x+1﹚]∴ lim (x→∞)[√﹙x²﹢x+1﹚]-[√﹙x²-x+1﹚]=lim (x→∞)2x/[√﹙x²﹢x+1﹚]+[√﹙x²-x+1﹚]=2 /2=1 展开全文阅读