设f是[0,1]上的连续函数,证明lim(n趋向于正无穷)n∫(从0到1)x^n...

Question

设f是[0,1]上的连续函数,证明lim(n趋向于正无穷)n∫(从0到1)x^nf(x)dx=f(1)

Answer 1

题目没有问题
∫{0,1}xⁿ*f(x)dx=∫{0,1-1/√n}xⁿ*f(x)dx+∫{1-1/√n,1}xⁿ*f(x)dx
由于f(x)在[0,1]上连续,xⁿ在[0,1]上不变号,且在[0,1]上可积
对f(x)在[0,1-1/√n]上运用积分第一中值定理,存在一点ξ₁∈[0,1-1/√n],使得
∫{0,1-1/√n}xⁿ*f(x)dx=f(ξ₁)*∫{0,1-1/√n}xⁿdx
=f(ξ₁)*[x^(n+1)/(n+1)]|
{0,1-1/√n}
=f(ξ₁)*(1-1/√n)^(n+1)/(n+1)
对f(x)在[1-1/√n,1]上运用积分第一中值定理,存在一点ξ₂∈[1-1/√n,1],使得
∫{1-1/√n,1}xⁿ*f(x)dx=f(ξ₂)*∫{1-1/√n,1}xⁿdx
=f(ξ₂)*[1/(n+1)-(1-1/√n)^(n+1)/(n+1)]
=f(ξ₂)/(n+1)[1-(1-1/√n)^(n+1)]
故lim{n→∞}n*∫{0,1}xⁿ*f(x)dx
=lim{n→∞}n*f(ξ₁)*(1-1/√n)^(n+1)/(n+1)+lim{n→∞}n*
f(ξ₂)/(n+1)[1-(1-1/√n)^(n+1)]
由于lim{n→∞}(1-1/√n)^(n+1)
=lim{n→∞}(1-1/√n)^n
=lim{x→0+}(1-x)^(1/x²)
=lim{x→0+}e^[1/x²*ln(1-x)]
=e^{lim{x→0+}[1/x²*(-x)]}
a→0,ln(1+a)~a
=0
故lim{n→∞}n*f(ξ₁)*(1-1/√n)^(n+1)/(n+1)
=lim{n→∞}f(ξ₁)*lim{n→∞}[n/(n+1)]*lim{n→∞}(1-1/√n)^(n+1)
=lim{n→∞}f(ξ₁)*1*0
=0
注：∵f(x)在[0,1]上连续,∴f(x)有界,∴lim{n→∞}f(ξ₁)为有限值
∵当n→∞时,1-1/√n→1,∴ξ₂→1
故lim{n→∞}
n*
f(ξ₂)/(n+1)[1-(1-1/√n)^(n+1)]
=lim{n→∞}f(ξ₂)*lim{n→∞}[n/(n+1)]*lim{n→∞}[1-(1-1/√n)^(n+1)]
=lim{ξ₂→1}f(ξ₂)*1*(1-0)
=f(1)
由连续性
因此,lim{n→∞}n*∫{0,1}xⁿ*f(x)dx=f(1),证毕
本题若直接根据积分中值定理,得到存在一点ξ∈(0,1),使得
∫{0,1}xⁿ*f(x)dx=ξⁿ*f(ξ),这里0