I want to proof the limit of the sequence $\{cos(a\_n)\}\_{n=0}^{\infty}$ where $a\_n \in \mathbb R$ and $lim\_{n \rightarrow \infty} a\_n = 0$ is equal to 1.

By intuition $lim\_{n \rightarrow \infty} cos(a\_n) = cos(lim\_{n \rightarrow \infty} a\_n)$, however I'm not familiar with any theorem allowing me to move the limit into a function.

I must prove: $|cos(a\_n) - 1| \le \epsilon$. Since $cos$ is monotonic decreasing on the interval $[0;\pi/2]$ my idea is:

Given $\epsilon > 0$. Suppose $\epsilon + 1 \in [0;1]$. Then choose $N \in \mathbb N$ such that $a\_n \le cos^{-1}(\epsilon + 1)$ for all $n \ge N$. Otherwise if $\epsilon + 1 \notin [0;1]$ then choose $N \in \mathbb N$ such that $a\_n \le cos^{-1}(1/2)$.

Advice is appreciated, thanks.