Chứng minh rằng :
$$\lim_{\mathit{x}\,\rightarrow \,\mathit{0}}\frac{\sin \mathit{x}}{\mathit{x}}= 1$$
$\sin \mathit{x}= \text{A}\left [ ...\,\left ( \mathit{x}+ \mathit{2}\,\pi \right )\left ( \mathit{x}+ \pi \right )x\left ( \mathit{x}- \pi \right )\left ( \mathit{x}- \mathit{2}\,\pi \right )\,... \right ]$
$\text{A}^{\,-\,\mathit{1}}= \frac{\pi }{\mathit{2}}\,{\prod\limits_{\mathit{k}= \mathit{1}}^{\infty }}\left [ \left ( \frac{\pi}{\mathit{2}} \right )^{\,\mathit{2}}- \mathit{k}^{\,\mathit{2}}\,\pi ^{\,\mathit{2}} \right ]$
Dùng nhân tử Wallis : $\lceil$ https://en.wikipedia.../Wallis_product $\rfloor$ , ta được :
$\lim_{\mathit{x}\,\rightarrow \,\mathit{0}}\frac{\sin \mathit{x}}{\mathit{x}}= \text{A}\left [ ...\,\left ( \mathit{2}\,\pi \right )\,\pi \,\left ( -\,\pi \right )\,\left ( -\,\mathit{2}\,\pi \right )\,... \right ]= \text{A}\,\prod\limits_{k= 1}^{\infty } \left ( -\,\mathit{k}^{\,\mathit{2}}\,\pi ^{\,\mathit{2}} \right )= \mathit{1}$