 Mathematics
How to prove the converse of lim (f(x) + g(x)) = limf(x) + lim g(x)

### Question Description

I don’t know how to handle this Calculus question and need guidance.

Prove the converse of the above using epsilon and delta?  I really have no clue Student has agreed that all tutoring, explanations, and answers provided by the tutor will be used to help in the learning process and in accordance with Studypool's honor code & terms of service. 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 limx->af(x) = L1 and limx->ag(x) = L2.

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)-(L1 + L2)|<ε.

Since |f(x)+g(x)-(L1 + L2)| < |f(x)-L1| + |g(x)-L2|,

we can chose a δ such that |f(x)-L1|< ϵ/2 and |g(x)-L2|< ϵ /2.

Since we know a δ1 exists such that if |x-a|<δ1 then |f(x)-L1|< ϵ /2 and another δ2 exists such that if |x-a|<δ2 then |g(x)-L2|< ϵ /2.

If we let δ be the minimum of δ1 and δ2.

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

QED Prashant S (1995)
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