How do we say the behavior of the limit of cosine square? > $\lim_{n\rightarrow \infty } (cosx)^{2n}$ Is it correct to say that this sequence diverges? > $\lim_{n\rightarrow \infty } ((cosx)^{2})^{n}$ > $ 0\le (cosx)^{2}\le1$, hence the behavior of $(cosx)^{2n}$ would be like oscillate between 0 and 1? But in some sense, I also feel like it's not continuous at $k\pi $ Thanks for helping me look into it!

The sequence of functions converges to a discontinuous function. For any $x$ such that $|\cos{x}| < 1$, the sequence tends to $0$, and otherwise (because of the $2$ in the power), it tends to $1$.

Hence, the limit function is simply

$$\lim_{n \to \infty} (\cos{x})^{2n} = \left\\{\begin{array}{lr} 1 &\text{ if } x = n\pi \\\ 0 & \text{ else} \end{array}\right.$$

where $n \in \mathbb{Z}$. This is continuous except at the points $n\pi$.