x→0のときのΓ(ax)sinπbxの極限

証明

$$\begin{array}{rl}\underset{x\to 0}{lim}\mathrm{\Gamma }(ax)\sin \pi bx& =\underset{x\to 0}{lim}\mathrm{\Gamma }(x)\sin \pi \frac{b}{a}x\\ & =\underset{x\to 0}{lim}x\mathrm{\Gamma }(x)\frac{\sin \pi \frac{b}{a}x}{x}\\ & =\underset{x\to 0}{lim}\mathrm{\Gamma }(x+1)\frac{\sin \pi \frac{b}{a}x}{x}\\ & =\pi \frac{b}{a}\underset{x\to 0}{lim}\mathrm{\Gamma }(x+1)\frac{\sin x}{x}\\ & =\frac{b}{a}\pi \end{array}$$