題目

Problem

Part III Solve the following problems

Given a family of parabolas $P_n:y=nx^2+\dfrac{1}{n}$ , where $n\in\mathbb{N}$ . Let $A_n$ be the area between $P_n$ and $P_{n+1}$ . Find the limit $\lim_{n\to\infty}\dfrac{A_n}{n^3}$ .

解答

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兩曲線之差為

P_n-P_{n+1} =&\,\left(nx^2+\frac1n\right)-\left((n+1)x^2+\frac1{n+1}\right) \\ =&\,\frac1{n(n+1)}-x^2.

交點由此式為 $0$ 得

x=\pm\frac1{\sqrt{n(n+1)}}.

在這兩點之間， $P_n-P_{n+1}\ge0$ ，故

\begin{align*} A_n =&\,\int_{-1/\sqrt{n(n+1)}}^{1/\sqrt{n(n+1)}} \left(\frac1{n(n+1)}-x^2\right)\,\mathrm{d}x \\ =&\,\frac4{3[n(n+1)]^{3/2}}. \end{align*}

因此

0\le\frac{A_n}{n^3} =&\,\frac4{3n^3[n(n+1)]^{3/2}} \longrightarrow0,

故 $\boxed{\displaystyle\lim_{n\to\infty}\frac{A_n}{n^3}=0}$ 。