# Limit of bounded functions

1. Oct 18, 2011

### Locoism

1. The problem statement, all variables and given/known data
Let f and g be real-valued functions defined on A ⊆ R and let c ∈ R be a cluster point of A. Suppose that f is bounded on a neighborhood of c and that limx→c g(x) = 0. Prove that limx→c f(x)g(x) = 0.

2. Relevant equations

3. The attempt at a solution
This isn't a very hard question, but it has to be done with no assumptions from calculus (Analysis 1).
Is it sufficient to say:
since f is bounded by (c-δ, c+δ) for some δ>0,
then limx→cf(x) = L is bounded by (f(c-δ), f(c+δ)),
and since f(x0) is a real number, for any x0 in that interval,

limx→c f(x)g(x) = limx→c f(x)limx→c g(x) = L * 0 = 0

I'm just not too sure what would be a formal proof...

2. Oct 18, 2011

### CompuChip

What do you mean with "if f is bounded by (a, b), then the limit is bounded by (f(a), f(b))"? Do you mean: if |f(x)| < a then |lim f(x)| < f(a)?
Because that is not necessarily true, if it even makes sense.

I think that the formal proof you are after uses the epsilon-delta definition, i.e.
$$\forall_{\epsilon > 0} \exists_{\delta(\epsilon) > 0} : |x - c| < \delta \implies |f(x)g(x)| < \epsilon$$
Of course you already know that
$$\forall_{\epsilon > 0} \exists_{\delta'(\epsilon) > 0} : |x - c| < \delta' \implies |g(x)| < \epsilon$$

3. Oct 18, 2011

### Locoism

Thank you, that makes sense