Calculus help

let the limit of f(x) as x->c = L
the limit of [f(x) + g(x)] as x->c does not exist
Assume that the limit of g(x) as x->c = M
Then lim(f) + lim(g) = L+M which belongs to the reals (or complex, depends on which field you are working on)
By the additive limit law lim (f(x)+g(x)) = lim(f)+lim(g) => lim(f)+lim(g) does not exist
but lim(f)+lim(g) = L+M which does exist ... (insert contradiction symbol here)
Hence, the limit of g cannot exist if lim(f+g) does not exist

proof by contradiction:

say that the limit of g(x) as x -> c did exist

assumptions: the limit of f(x) as x -> c exists
limit of [f(x) + g(x)] as x -> c does not exist

Let eps > 0, the hypothetical limit of g(x) as x -> c as L and the limit of f(x) as x-> c as W.

then there for every epsilon > 0, there exists a delta > 0 such that if | x - c | < delta then | g(x) - L | < eps/2.

and

for every epsilon > 0, there exists a delta > 0 such that if | x - c | < delta then | f(x) - W | < eps/2.

But we know from the triangle inequality that | f(x) - W + g(x) - L | <= | f(x) - W | + | g(x) - L| < eps and so, | [ f(x) + g(x) ] - [ W + L ] | < eps

but then this implies that:

for every epsilon > 0, there exists a delta > 0 such that if | x - c | < delta then | f(x) + g(x) - (W + L) | < eps, or that the limit of f(x) + g(x) is W+L, which is a contradiction from our assumptions.

really, really shitty notation and poor proof-writing skills. I blame the summer vacation for doing that, not to mention that I never really learn how to write proofs well. Hell, for all I know I missed something big. But that's a quick outline of my proof.

Let epsilon = e > 0
There exists an L and a delta = d1 such that if 0 < | x - c | < d1 then | f(x) - L | < e where L belongs to the reals.
And there exists an M and a delta = d2 such that if 0 < | x - c | < d2 then | g(x) - M | < e where M belongs to the reals, M =/= 0.

Consider f(x)/g(x). Choose d = min{d1, d2} (then clearly f converges to L and g converges to M for |x - c | < d). Assume 0 < | x - c | < d, define h(x) = 1/g(x) => lim(h(x)) = 1/M = m (since M =/= 0, the proof is easy, let me know if you want it). So, |f(x)*h(x) - Lm| = |(f(x)*h(x)-f(x)*m) + (f(x)*m - L*m)| =< |f(x)|*|(h(x) - m)|+|m|*|(f(x) - L)|. Since both h(x) and f(x) have a limit, they are bounded on the delta neighborhood of c. Hence, there exists and number Z such that |f(x)|*|(h(x) - m)|+|m|*|(f(x) - L)| <= Z*|h(x) - m| + Z*|f(x) - L|. Since e arbitrary, we can easily find conditions for which for 0 < | x - c | < d => |f(x) - L| < e/(2*Z) and |h(x) - m| < e/(2*Z). So we arrive at the result;
Z*|h(x) - m| + Z*|f(x) - L| =< Z*e/(2*Z) + Z*e/(2*Z) = e.
And we are done

Messy, but it holds (I hate writing proofs out on the computer)