Prove $\lim_\limits{n \to \infty}-\cos(\pi\sqrt{4n^2+10})$ exists and is equal to $-1$ My idea is to take two subsequence converging to different limits. On this step I have problem. How can I set equation  of subsequence when $\cos$ has certain value and $n \to \infty$?
$$
\lim_{n \to \infty}-\cos(\pi\sqrt{4n^2+10})
$$
If I say that $\frac{5}{4n^2} \to 0$ than $n \to \infty$, I have $\lim_{n \to \infty}-\cos(2\pi n) = -1$. Is this right?

One may also argue that $$\cos(\pi\sqrt {4n^2+10})=\cos(\pi\sqrt {4n^2+10}-2\pi n)  = \cos\left(\pi\frac{(\sqrt {4n^2+10})^2-4n^2}{\sqrt {4n^2+10}+2n}\right)  = \cos\left(\pi\frac{10}{\sqrt {4n^2+10}+2n}\right).$$
Clearly, $\lim_{n\to\infty}\frac{10}{\sqrt {4n^2+10}+2n}=0$, hence, by continuity,
$$\lim_{n\to\infty} \cos(\pi\sqrt {4n^2+10})=\lim_{n\to\infty}\cos\left(\pi\frac{10}{\sqrt {4n^2+10}+2n}\right)=\cos 0=1. $$
