# Thread: Limit Problem pt. 2

1. ## Limit Problem pt. 2

We'll say a function f unhelpful at a if, whenever g is a function such that lim x goes to a of g (x) does not exist, then lim x goes to a of (f(x)+g(x)) also does not exist. Prove that a function f is unhelpful at a if an only if lim x goes to a of f(x) does exist.

I'm having trouble reading this problem. Can anyone phrase it in another way that might make more sense?

Right now I understand the for f(X) to be considered an unhelpful function

(1) the limit of f(x) as x goes to a must exist.

(2) the limit of g(x) as x goes to a does not exist

(3) the limit of (f(x)+g(x)) as x goes to a does not exist

In condition 3, if we know that the limit of the sum is equal to the sum of the limits, then we know that the limit of f(x) as x goes to a and the limit of g(x) as x goes to a must both not exist. How can that be if we also know in condition 1 that the limit of f(x) as x goes to a must exist?

2. ## Re: Limit Problem pt. 2

It would probably help to write the definition symbolically. If f is any function, let L(f) mean that $\lim_{x\to a}f(x)$ exists. I'll write ~L(f) to mean the negation of L(f), i.e., that $\lim_{x\to a}f(x)$ does not exist. The definition says that f is unhelpful if

∀g. (~L(g) => ~L(f + g)).

Note that the definition quantifies over all functions g, so the formula above is a property of f only. Thus, when we say that f is unhelpful, there is no g to talk about. This is analogous to saying that x is the least element of a set S if ∀y ∈ S. x <= y. Again there is quantification over y, so this is a property of x only. When we say that x is the least element of S, we don't have any y in the context of our discourse.

The theorem says that for any function f, if f is unhelpful, then L(f). One way to prove it it to use the facts that L(f) implies L(-f) for all f and that L(0). These are the only fact about L that are needed, the rest of the proof is simply a logical manipulation.