求极限和间断点?
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求极限和间断点?
![](http://img.wesiedu.com/upload/2/81/281eb43ec0b6bdad19c01515c4873083.jpg)
![](http://img.wesiedu.com/upload/2/81/281eb43ec0b6bdad19c01515c4873083.jpg)
![求极限和间断点?](/uploads/image/z/15108309-45-9.jpg?t=%E6%B1%82%E6%9E%81%E9%99%90%E5%92%8C%E9%97%B4%E6%96%AD%E7%82%B9%3F%26nbsp%3B)
解 要使
0 = lim(x→+∞)[√(x²-x+1) - (β-αx)]
= lim(x→+∞)[(x²-x+1) - (β-αx)²]/[√(x²-x+1) + (β-αx)]
= lim(x→+∞)[(1-α²)x²+(2αβ-1)x+(1-β²)]/[√(x²-x+1) + (β-αx)]
= lim(x→+∞)[(1-α²)x+(2αβ-1)+(1-β²)/x]/[√(1-1/x+1/x²) + (β/x-α)]
成立,须有
1-α² = 0,2αβ-1 = 0,1-α≠0,
由此可解得
α = -1,β = -1/2,
故选 C.
0 = lim(x→+∞)[√(x²-x+1) - (β-αx)]
= lim(x→+∞)[(x²-x+1) - (β-αx)²]/[√(x²-x+1) + (β-αx)]
= lim(x→+∞)[(1-α²)x²+(2αβ-1)x+(1-β²)]/[√(x²-x+1) + (β-αx)]
= lim(x→+∞)[(1-α²)x+(2αβ-1)+(1-β²)/x]/[√(1-1/x+1/x²) + (β/x-α)]
成立,须有
1-α² = 0,2αβ-1 = 0,1-α≠0,
由此可解得
α = -1,β = -1/2,
故选 C.