求limn→∞(1/1×2+1/2×3+…+1/n×n+1)和若limχ→1[(χ^2+aχ+b)÷(1-χ)]=4,求ab的值?要
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求limn→∞(1/1×2+1/2×3+…+1/n×n+1)和若limχ→1[(χ^2+aχ+b)÷(1-χ)]=4,求ab的值?要
求limn→∞(1/1×2+1/2×3+…+1/n×n+1)和若limχ→1[(χ^2+aχ+b)÷(1-χ)]=4,求ab的值?要
求limn→∞(1/1×2+1/2×3+…+1/n×n+1)和若limχ→1[(χ^2+aχ+b)÷(1-χ)]=4,求ab的值?要
1/1×2 + 1/2×3+…+1/n×n+1
=(1 - 1/2)+(1/2 -1/3)+...+(1/n -1/(n+1))
=1-1/(n+1)
limn→∞(1/1×2+1/2×3+…+1/n×n+1)=limn→∞(1-1/(n+1))=1
limχ→1[(χ^2+aχ+b)÷(1-χ)]的极限存在(=4),
分母->0,所以分子必->0,
可知limχ→1(χ^2+aχ+b)=0,
所以b=0,ab就是0.
事实上,a也可以求的:
代入h=1-χ,
limχ→1[(χ^2+aχ+b)/(1-χ)]
=limχ→1[(1-h)^2+a(1-h)+b]/h
=d[(1-x)^2+a(1-x)+b]/dx |x=0
=[-2(1-x)-a]|x=0
=-2-a
所以-2-a=4
a=-6
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