Confused about limit proofs conceptually In a question like this:
Prove that if $\lim_{x \to a} f(x) = l$ and $\lim_{x \to a} g(x) = m$ then $\lim_{x \to a} \max(f(x), g(x)) = \max(l, m)$ 
In general, when asked for proofs like this, are we supposed to find a $\delta$ or is the presupposition that the limit already exists, so that it is already true that $|h(x)−Q|<\epsilon$ for $|x−a|<\delta′$??Or do we first prove that there exists a $\delta$? But how can we do that if all we are given is a GENERAL $f(x), g(x)$?
Thanks!
I am confused because I am not sure how to prove things then. 
 A: $\textbf{Hint:}$
for any real numbers $x, y$ and $a, b$
$$|\max(x, y) - \max(a, b)|\le \max(|x-a|,|y-b|)$$
A: The first thing to do in proofs like this is to find a good $\epsilon$. We might as well suppose $m > l$. Now if $\epsilon$ is small enough, $m-\epsilon > l+\epsilon$. 
Now for this small choice of $\epsilon$, find $\delta > 0$ so that whenever $|x_0-x| < \delta$, we have both $|f(x)-f(x_0)| < \epsilon$ and $|g(x)-g(x_0)|<\epsilon$ (why can we do this?). Since $\epsilon$ was "small," we also know that for any such $x_0$, we have $max(f(x_0), g(x_0)) = g(x_0) \in (m-\epsilon, m+\epsilon)$. But this is exactly what we wanted to show!
A: let $h=max\{f,g\}$ . let $k=max\{l,m\} $.
Choose  a common $\delta>0$ for both $ f,g$ such that $0<|x-a|<\delta$ $\implies l-\epsilon <f(x)<l+\epsilon $ and $ m-\epsilon <g(x)<m+\epsilon $  
Now $f(x)<l+ \epsilon<k+\epsilon$ and  $g(x)<m+ \epsilon<k+\epsilon$ 
So $h(x)<k+\epsilon $
Again $l-\epsilon <f(x)$ and $m-\epsilon <g(x)$
So $h(x)>max\{l-\epsilon,m-\epsilon\}=max\{l,m\}-\epsilon =k-\epsilon$
hence proved
