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?

HINT:
Let
$$
a_n=\sqrt{4n^2+10}-2n={10\over\sqrt{4n^2+10}+2n}.
$$
It is clear that $\lim\limits_{n\to\infty}a_n=0$, and:
$$
\cos(\pi\sqrt{4n^2+10})=\cos(2n\pi+\pi a_n)=\cos(\pi a_n).
$$
