real analysis - Sequence defined by max$left{a_n, b_n right}$. Proving convergence

Assume $(a_n)$ and $(b_n)$ are two real sequences, and define $$c_n = \text{max}\left\{a_n, b_n\right\}$$ for $n \in \mathbb{N}$. Suppose $(a_n)$ and $(b_n)$ are two convergent sequences. Prove then that $(c_n)$ is also a convergent sequence, and that $$\lim_{n \to \infty} c_n = \text{max} \left\{ \lim_{n \to \infty} a_n, \lim_{n \to \infty} b_n \right\}.$$

Attempt: $(a_n)$ and $(b_n)$ are both convergent. Hence there exists an $n_0 \in \mathbb{N}$ such that $\forall n \geq n_0: | a_n - L | < \epsilon$. Furthermore, there exists an $n_1 \in \mathbb{N}$ such that $\forall n \geq n_1: | b_n - K | < \epsilon$. Here $L$ and $K$ are the limits of resp. $(a_n)$ and $(b_n)$. Now let $n_2 = \text{max} \left\{n_0, n_1 \right\}$. Let $n \geq n_2$ be arbitrary. Then we have that $L - \epsilon < a_n < L + \epsilon$ and $K - \epsilon < b_n < K + \epsilon$. Since $c_n = \text{max}\left\{a_n, b_n\right\}$, we have that $c_n \geq a_n$ and $c_n \geq b_n$. So $L - \epsilon < a_n \leq c_n$ and thus $L - \epsilon < c_n$.

But now I'm stuck. Help would be appreciated!

Answer

Remember that $\max\{x,y\}=\frac{x+y+|x-y|}{2}$.

Thus $c_n=\frac{a_n+b_n+|a_n-b_n|}{2}$. So, if $\lim a_n=a$ and $\lim b_n=b$:

$$\lim c_n=\frac{1}{2}(\lim a_n+\lim b_n+|\lim a_n-\lim b_n|)=\frac{1}{2}(a+b+|a-b|)=\max\{a,b\}$$