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