$\displaystyle{\lim_{n \to \infty} {\rm sinc}^{n} \left(\frac{x}{\sqrt{n}} \right) = e^{-\frac{x^2}{6}}}$ の証明
\begin{eqnarray} {\rm sinc}^n \left(\frac{x}{\sqrt{n}} \right) &=& \left(\frac{\sin \frac{x}{\sqrt{n}}}{\frac{x}{\sqrt{n}}} \right)^n \\ &=& \left\{\frac{\sqrt{n}}{x} \left\{\frac{x}{\sqrt{n}} - \frac{1}{6} \left(\frac{x}{\sqrt{n}} \right)^3 +O\left(\left(\frac{x}{\sqrt{n}} \right)^5 \right) \right\} \right\}^n \\ &=& \left\{1 - \frac{1}{6} \left(\frac{x}{\sqrt{n}} \right)^2 +O\left(\left(\frac{x}{\sqrt{n}} \right)^4 \right) \right\}^n \\ &=& \left\{1 - \frac{1}{6} \frac{x^2}{n} +O\left(\left(\frac{x^2}{n} \right)^2 \right) \right\}^n \\ &=& 1 - \frac{x^2}{6} +O\left( \left(\frac{x^2}{n} \right)^2 \right) \\ &\to & e^{-\frac{x^2}{6}} \ \ \ (n \to \infty) \end{eqnarray}
ランダウのO記法と二項定理によるｎ乗の展開を用いている．
参考サイト
wlframalpha https://www.wolframalpha.com/input/?i=(sinc(x%2F%E2%88%9A(n)))%5En
