lim (x→0)[√﹙x²﹢x+1﹚]-[√﹙x²-x+1﹚]求极限
问题描述:
lim (x→0)[√﹙x²﹢x+1﹚]-[√﹙x²-x+1﹚]求极限
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