limit laws:$\lim_{n\to\infty}\max(a_n,b_n)=\max(\lim_{n\to\infty}a_n,\lim_{n\to\infty}b_n)$
Assume that $a_n \to a$ and $b_n \to b$.
The most simple is to split this in two cases.
- if $a\neq b$: Assume without loss of generality that $a<b$. Let $\epsilon = \frac {b-a}2$. You can find $N_{1,2}$ such as \begin{align} n > N_1 &\implies |a_n - a|<\epsilon\\ n > N_2 &\implies |b_n - b|<\epsilon\\ \implies [n\ge \max(N_1, N_2) &\implies |a_n - a|<\epsilon, |b_n - b|<\epsilon] \end{align} Now if $n\ge \max(N_1, N_2)$: $$a_n<a+\epsilon=b-\epsilon<b_n\\ $$ so $\max (a_n, b_n) = b_n \to b = \max (a,b)$.
- If $a=b$: let $\epsilon>0$. You can find $N_{1,2}$ such as \begin{align} n > N_1 &\implies |a_n - a|<\epsilon\\ n > N_2 &\implies |b_n - a|<\epsilon\\ \implies [n\ge \max(N_1, N_2) &\implies |a_n - a|<\epsilon, |b_n - a|<\epsilon]\\ \implies [n\ge \max(N_1, N_2) &\implies |\max (a_n,b_n) - a|<\epsilon]\\ \end{align} so $\max (a_n,b_n)\to a = \max(a,b)$.
Separate your task into two cases:
- If $a=\lim_{n\to\infty} a_n \neq \lim_{n\to\infty} b_n=b$, then you can assume, without loss of generality, that $a>b$. In this case, you can show that from some $n$ on, $a_n > b_n$ and thus that $\lim_{n\to\infty}\max(a_n,b_n) = \lim_{n\to\infty} a_n = a$
- If $a=b$, then it is best to go to the definition of the limit using epsilons to prove that $a$ must be the limit of the max sequence.