Prove that a sequence is convergent and determine its limit 
I have to prove that the sequence $(x_n)_{n \in \mathbb R}$ defined by:
  $x_n=\cos^2(\pi \sqrt{(n+1)(n+3})$ is convergent and determine its limit.

If I find that the limit exists (and its finite) doesn' t that mean that the sequence is convergent? 
I wrote $x_n=1-\sin^2(\pi \sqrt{(n+1)(n+3})$ and for $n\rightarrow \infty$, $\sin^2(\pi \sqrt{(n+1)(n+3}) \sim \sin^2(\pi \sqrt{n^2})=\sin^2(\pi n)=0$. So the limit is actually $1$. Is this correct?

Note that $(n+1)(n+3)=(n+2)^2-1$.
Also, $$\left((n+2)-\frac1{2(n+2)}\right)^2=(n+2)^2-1+\frac1{4(n+2)^2}$$ is slightly bigger than $(n+2)^2-1$, whereas
$$\left((n+2)-\frac1{n+2}\right)^2=(n+2)^2-2+\frac1{(n+2)^2}$$ is strictly smaller than $(n+2)^2-1$ (because $\frac1{(n+2)^2}<1$).
We conclude
$$ (n+2)-\frac1{n+2}<\sqrt{(n+1)(n+3)}<(n+2)-\frac1{2(n+2)}$$
And as $\cos (\pi a)=\pm \cos(\pi(a-n-2))$, we conclude that 
$x_n=\cos^2 (\pi\delta_n)$ where $-\frac1{n+2}<\delta_n<-\frac1{2(n+2)}$. We see $\delta_n\to 0$ and hence $x_n\to\cos^20=1$.
