作业帮求极限和间断点?
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/23 17:38:53
求极限和间断点?
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.