limit proof

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September 20th 2008, 12:14 PM
PvtBillPilgrim

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?
September 20th 2008, 12:24 PM
hwhelper

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.
September 20th 2008, 12:32 PM
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?
September 20th 2008, 12:57 PM
Jhevon

Quote:

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 $\epsilon > 0$ and all those $\delta$ 's and stuff :p that

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

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

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

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

i don't really like this though. hwhelper's way is perhaps better. maybe you can fix it up
September 20th 2008, 01:09 PM
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?
September 20th 2008, 01:25 PM
Jhevon

Quote:

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
September 20th 2008, 05:22 PM
ThePerfectHacker

Quote:

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.