[tex]\text{Functia sinus e o functie periodica cu perioada 2k}\pi,k\in \mathbb{Z} .\\\text{Asadar:}\\a_n= \sin(\pi \sqrt{4n^2+n-1} +2k\pi}).\\\text{Pentru k=- n obtinem:}\\a_n= \sin(\pi \sqrt{4n^2+n-1} -2\cdot n\cdot \pi})\\\displaystyle \limit\lim_{n\to\infty} a_n=\lim_{n\to \infty} \sin (\pi (\sqrt{4n^2+n-1}-2n)) =\\\lim_{n\to \infty} \sin \left(\pi\cdot \dfrac{4n^2+n-1- 4n^2}{\sqrt{4n^2+n-1}+2n}\right) =\lim_{n\to\infty } \sin \left (\pi \dfrac{n-1}{\sqrt{4n^2+n-1}+2n}\right)=\\=\sin \dfrac{\pi}{4} =\dfrac{\sqrt{2}}{2} [/tex]
