lim (x→0)[√﹙x²﹢x+1﹚]-[√﹙x²-x+1﹚]求极限
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lim (x→0)[√﹙x²﹢x+1﹚]-[√﹙x²-x+1﹚]求极限
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
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
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