How to Prove the Limit of a Product of Functions?

[Locoism]

Homework Statement

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 lim_x→c g(x) = 0. Prove that lim_x→c f(x)g(x) = 0.

Homework Equations

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 lim_x→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,

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

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

 

[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 &gt; 0} \exists_{\delta(\epsilon) &gt; 0} : |x - c| &lt; \delta \implies |f(x)g(x)| &lt; \epsilon
Of course you already know that
\forall_{\epsilon &gt; 0} \exists_{\delta&#039;(\epsilon) &gt; 0} : |x - c| &lt; \delta&#039; \implies |g(x)| &lt; \epsilon

 

[Locoism]

Thank you, that makes sense