We define for all real $x > 0$ the sequence $(I_{n}(x))_{n \geqslant 1}$ by: $$I_{n}(x) = \int_{0}^{n} \left(1 - \frac{t}{n}\right)^{n} t^{x-1} dt$$ Show that, for all $x > 0$, $$\lim_{n \rightarrow +\infty} I_{n}(x) = \Gamma(x)$$
We define for all real $x > 0$ the sequence $(I_{n}(x))_{n \geqslant 1}$ by:
$$I_{n}(x) = \int_{0}^{n} \left(1 - \frac{t}{n}\right)^{n} t^{x-1} dt$$
Show that, for all $x > 0$,
$$\lim_{n \rightarrow +\infty} I_{n}(x) = \Gamma(x)$$