How to Prove the Limit of a Product of Functions?

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SUMMARY

The discussion focuses on proving that if \( f \) is a bounded function near a cluster point \( c \) and \( \lim_{x \to c} g(x) = 0 \), then \( \lim_{x \to c} f(x)g(x) = 0 \). The proof utilizes the epsilon-delta definition of limits, establishing that for any \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that \( |f(x)g(x)| < \epsilon \) when \( |x - c| < \delta \). The participants clarify the necessity of formal proof and the implications of boundedness in the context of limits.

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

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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...
 
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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.
[tex]\forall_{\epsilon > 0} \exists_{\delta(\epsilon) > 0} : |x - c| < \delta \implies |f(x)g(x)| < \epsilon[/tex]
Of course you already know that
[tex]\forall_{\epsilon > 0} \exists_{\delta'(\epsilon) > 0} : |x - c| < \delta' \implies |g(x)| < \epsilon[/tex]
 
Thank you, that makes sense
 

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