Question

Mă puteți ajuta,va rog mult? AM213.

Answer 1

$I(n) = \displaystyle \int_{0}^1(2017+x^{n})^n\, dx \\ \\ x\in(0,1) \Rightarrow x^n \to 0,\quad \text{deoarece atunci cand }n\to \infty,~x^n\text{ devine neglijabil}\\ \\ I(n)\Big|_{n\to \infty} = \int_{0}^1(2017+0)^n\, dx = \int_{0}^12017^n\, dx = \\ \\ =2017^n x\Big|_{0}^1 = 2017^n \\ \\ \Rightarrow \sqrt[n]{I(n)}\Big|_{n\to \infty} = \sqrt[n]{2017^n} = \boxed{2017}$