# Thread: limit proof

1. ## limit proof

I have that the limit of f(x) as x goes to c exists and that the limit of g(x) as x goes to c does not exist.

How do I prove that the limit of h(x) as x goes to c does not exist (where h = f + g)?

What can you say about the limit of g using epsilon notation to reach a contradiction?

2. lim g(x) = lim h(x) - lim f(x)
because the "limit" operation distributes over addition/subtraction. <-- Are you allowed to assume this in your problem or do you essentially have to prove this?
If RHS is finite then LHS is finite. QED.

3. I guess that does work, but I don't want to use that.

abs(g(x) - L) >= epsilon I think, but g doesn't technically have a limit, so what do you say?

4. Originally Posted by PvtBillPilgrim
I guess that does work, but I don't want to use that.

abs(g(x) - L) >= epsilon I think, but g doesn't technically have a limit, so what do you say?
i guess that could work.

you have for all $\displaystyle \epsilon > 0$ and all those $\displaystyle \delta$'s and stuff that

$\displaystyle |f(x) - F| < \frac {\epsilon}2$ but $\displaystyle |g(x) - G| \ge \frac {\epsilon}2 = \frac {\epsilon}2 + \zeta$, where $\displaystyle \zeta > 0$

where $\displaystyle F$ and $\displaystyle G$ are the limits of $\displaystyle f(x)$ and $\displaystyle g(x)$ respectfully.

had $\displaystyle \lim_{x \to c}h(x)$ existed, we would have $\displaystyle |h(x) - H| < \epsilon$, where $\displaystyle H = F + G$ is the limit of $\displaystyle h(x)$

but, $\displaystyle |h(x) - H| = |[f(x) + g(x)] - [F + G]| = |[f(x) - F] + [g(x) - G]|$ $\displaystyle \le |f(x) - F| + |g(x) - G| = \frac {\epsilon}2 + \bigg( \frac {\epsilon}2 + \zeta \bigg) > \epsilon$

i don't really like this though. hwhelper's way is perhaps better. maybe you can fix it up

5. it's weird because at the end you get that

abs(f - F) < epsilon
and abs(g- G) >= epsilon

does this imply that abs(h - (F+G)) >= epsilon?

6. Originally Posted by PvtBillPilgrim
it's weird because at the end you get that

abs(f - F) < epsilon
and abs(g- G) >= epsilon

does this imply that abs(h - (F+G)) >= epsilon?
yeah, the inequalities turned in opposing directions does make things unclear. that's why i liked hwhelper's solution better. it is more straight forward. an epsilon proof is more subtle, i'll have to think about it more

7. Originally Posted by PvtBillPilgrim
I have that the limit of f(x) as x goes to c exists and that the limit of g(x) as x goes to c does not exist.

How do I prove that the limit of h(x) as x goes to c does not exist (where h = f + g)?

What can you say about the limit of g using epsilon notation to reach a contradiction?
If the limit of (f+g) existed then the limit of [(f+g)-f] = g would have existed.
A contradiction.