$\lim_{n\to +\infty} |(2n\pi+1/n) \sin(2n\pi+1/n)|$, $n\in \mathbb N$. I tried L'Hospital which gives $$ \lim_{n\to +\infty}\left| \frac{\sin(2n\pi+1/n)}{(2n\pi+1/n)^{-1}} \right| = \lim_{n\to +\infty}\left| \frac{\cos(\frac{2\pi x^2+1}{x})(2\pi x^2+1)^2}{x^2} \right|.$$
Is the limit $+\infty$? I am not sure because of the cosine term.
Note that $\sin$ has period $2\pi$, so $\sin(2n\pi+\frac 1 n)=\sin(\frac 1 n)$. $\lim (|2n\pi+\frac 1 n)\sin(\frac 1 n)|=\lim (|2n\pi+\frac 1 n)(\frac 1 n)|=2\pi$. I have used the fact that $\frac {\sin\, x } x \to 1$ as $ x \to 0$.