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

Calculus
Tutor: None Selected Time limit: 1 Day

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

Aug 2nd, 2014

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

Aug 2nd, 2014

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