[数分提高]2014-2015-2第7教学周第2次课 (2015-04-16)

1. 设 $0<f\in C[0,1]$, 试求 $$\bex \vlm{n}\sqrt[n]{f\sex{\frac{1}{n}}f\sex{\frac{2}{n}}\cdots f\sex{\frac{n-1}{n}}f(1)}. \eex$$

 

解答: $$\bex \mbox{原极限}=\exp\sez{\vlm{n}\frac{1}{n}\sum_{k=1}^n \ln f\sex{\frac{k}{n}}} =\exp\sez{\int_0^1 \ln f(x)\rd x}. \eex$$

 

2. 设 $$\bex f(x)=\int_x^{x^2} \sex{1+\frac{1}{2t}}^t \sin \frac{1}{\sqrt{t}}\rd t,\quad x>0. \eex$$ 试求 $$\bex \vlm{n}f(n)\sin\frac{1}{n}. \eex$$

 

解答: 由 $$\bex \vlm{t}\sex{1+\frac{1}{2t}}^t =\vlm{t}\sez{\sex{1+\frac{1}{2t}}^{2t}}^\frac{1}{2} =e^\frac{1}{2}\quad\sex{t\to\infty}, \eex$$ $$\bex \vlm{t}\frac{\sin\frac{1}{\sqrt{t}}}{\frac{1}{\sqrt{t}}} =\lim_{x\to 0}\frac{\sin x}{x} =1 \eex$$ 知 $\forall\ \ve>0$, $\exists\ X>0,\st$ $$\bex t\geq X\ra e^\frac{1}{2}-\ve<\sex{1+\frac{1}{2t}}^t <e^\frac{1}{2}+\ve,\quad \frac{1-\ve}{\sqrt{t}} <\sin\frac{1}{\sqrt{t}}<\frac{1+\ve}{\sqrt{t}}. \eex$$ 于是当 $x\geq X$ 时, $$\beex \bea f(x)&=\int_x^{x^2} \sex{1+\frac{1}{2t}}^t\sez{e^{\frac{1}{\sqrt{t}}}-1}\rd t\\ &\geq \sex{e^\frac{1}{2}-\ve}(1-\ve)\int_x^{x^2}\frac{1}{\sqrt{t}}\rd t\\ &=2(e^\frac{1}{2}-\ve)(1-\ve) (x-\sqrt{x}),\\ f(x)&\leq 2(e^\frac{1}{2}+\ve)(1+\ve) (x-\sqrt{x}). \eea \eeex$$ 综上上两式, $$\bex 2(e^\frac{1}{2}-\ve)(1-\ve)\sex{1-\frac{1}{\sqrt{x}}} \leq\frac{f(x)}{x}\leq 2(e^\frac{1}{2}+\ve)(1+\ve)\sex{1-\frac{1}{\sqrt{x}}}. \eex$$ 令 $x\to\infty$, 有 $$\bex 2(e^\frac{1}{2}-\ve)(1-\ve)\leq \vli{x}\frac{f(x)}{x} \leq \vls{x}\frac{f(x)}{x} \leq 2(e^\frac{1}{2}+\ve)(1+\ve). \eex$$ 再令 $\ve\to0$ 有 $$\bex \vlm{x}\frac{f(x)}{x}=2e^\frac{1}{2}, \eex$$ $$\bex \vlm{n}f(n)\sin\frac{1}{n} =\vlm{n}\frac{f(n)}{n}=2\sqrt{e}. \eex$$