Convergence of $\cos(nx)$ for $x\neq k \pi$

Question

I've found that $cos(nx) =-\frac{(sin((n+1)x)-sin(nx)cos(x))}{(sinx))}.$

My question here is how I can show that this succession is convergent for n approaching $ \to \infty $

Answer 1

If $\cos(nx)$ converges to a limit $c$ , the subsequence $\cos(2nx)$ also converges to $c$. But $\cos(2nx)=2\cos(nx)^2-1$, so $c=2c^2-1$. In particular, $c\neq 0$.

Now $\cos((n+1)x)+\cos((n-1)x)=2\cos(nx)\cos(x)$. Taking limits shows that $2c=2c\cos(x)$. Since $c\neq 0$, this implies $\cos(x)=1$, which is not possible since $x\neq k\pi,k\in\mathbb{Z}$. Contradiction.