1. 设 $0<fin C[0,1]$, 试求 $$ex vlm{n}sqrt[n]{fsex{frac{1}{n}}fsex{frac{2}{n}}cdots fsex{frac{n-1}{n}}f(1)}. eex$$
解答: $$ex mbox{原极限}=expsez{vlm{n}frac{1}{n}sum_{k=1}^n ln fsex{frac{k}{n}}} =expsez{int_0^1 ln f(x) d x}. eex$$
2. 设 $$ex f(x)=int_x^{x^2} sex{1+frac{1}{2t}}^t sin frac{1}{sqrt{t}} d t,quad x>0. eex$$ 试求 $$ex vlm{n}f(n)sinfrac{1}{n}. eex$$
解答: 由 $$ex vlm{t}sex{1+frac{1}{2t}}^t =vlm{t}sez{sex{1+frac{1}{2t}}^{2t}}^frac{1}{2} =e^frac{1}{2}quadsex{t oinfty}, eex$$ $$ex vlm{t}frac{sinfrac{1}{sqrt{t}}}{frac{1}{sqrt{t}}} =lim_{x o 0}frac{sin x}{x} =1 eex$$ 知 $forall ve>0$, $exists X>0,st$ $$ex tgeq X a e^frac{1}{2}-ve<sex{1+frac{1}{2t}}^t <e^frac{1}{2}+ve,quad frac{1-ve}{sqrt{t}} <sinfrac{1}{sqrt{t}}<frac{1+ve}{sqrt{t}}. eex$$ 于是当 $xgeq X$ 时, $$eex ea f(x)&=int_x^{x^2} sex{1+frac{1}{2t}}^tsez{e^{frac{1}{sqrt{t}}}-1} d t\ &geq sex{e^frac{1}{2}-ve}(1-ve)int_x^{x^2}frac{1}{sqrt{t}} d t\ &=2(e^frac{1}{2}-ve)(1-ve) (x-sqrt{x}),\ f(x)&leq 2(e^frac{1}{2}+ve)(1+ve) (x-sqrt{x}). eea eeex$$ 综上上两式, $$ex 2(e^frac{1}{2}-ve)(1-ve)sex{1-frac{1}{sqrt{x}}} leqfrac{f(x)}{x}leq 2(e^frac{1}{2}+ve)(1+ve)sex{1-frac{1}{sqrt{x}}}. eex$$ 令 $x oinfty$, 有 $$ex 2(e^frac{1}{2}-ve)(1-ve)leq vli{x}frac{f(x)}{x} leq vls{x}frac{f(x)}{x} leq 2(e^frac{1}{2}+ve)(1+ve). eex$$ 再令 $ve o0$ 有 $$ex vlm{x}frac{f(x)}{x}=2e^frac{1}{2}, eex$$ $$ex vlm{n}f(n)sinfrac{1}{n} =vlm{n}frac{f(n)}{n}=2sqrt{e}. eex$$