### Question Description

Prove the converse of the above using epsilon and delta? I really have no clue

## Explanation & Answer

Disclaimer: I did this stuff a while back, so I don't know if this over what you are supposed to know or at the right level.

The converse of this statement would be: lim f(x) + lim g(x) = lim (f(x)+g(x))

Assume lim_{x->a}f(x) = L_{1} and lim_{x->a}g(x)
= L_{2}.

We need to show that for any ε>0 there will exist a δ such that
if |x-a|<δ then |f(x)+g(x)-(L_{1} + L_{2})|<ε.

Since |f(x)+g(x)-(L_{1} + L_{2})| <
|f(x)-L_{1}| + |g(x)-L_{2}|,

we can chose a δ such that |f(x)-L_{1}|< ϵ/2 and
|g(x)-L_{2}|< ϵ /2.

_{1}exists such that if |x-a|<δ

_{1}then |f(x)-L

_{1}|< ϵ /2 and another δ

_{2}exists such that if |x-a|<δ

_{2}then |g(x)-L

_{2}|< ϵ /2.

If we let δ
be the minimum of δ_{1} and δ_{2}.

We see that: lim f(x) + lim g(x) = lim (f(x)+g(x))

QED