How does f(x)g(x)->0 if f(x)->0 and g(x) is limited?

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Discussion Overview

The discussion centers around the limit of the product of two functions, specifically exploring the condition under which \( f(x)g(x) \to 0 \) if \( f(x) \to 0 \) and \( g(x) \) is limited. The scope includes theoretical aspects of calculus and limit proofs.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant requests a proof for the limit product rule, indicating difficulty in finding it in various calculus books.
  • Another participant provides a link to a proof of the product rule for limits, suggesting it may be relevant.
  • A participant expresses confusion about the epsilon-delta definition of limits, questioning whether they need to show \( f(x)g(x) < \epsilon \) given \( \epsilon > 0 \).
  • There is a request for clarification on what is meant by \( g(x) \) being "limited," with one participant suggesting it means \( g(x) \) does not approach infinity and provides an example involving the sine function.
  • A participant distinguishes between bounded and convergent sequences, noting that a bounded sequence may not converge, using the sequence \( a_n = (-1)^n \) as an example.
  • Another participant shares a proof from their notes, expressing difficulty in understanding a specific part regarding \( |g(x)| \leq B \).

Areas of Agreement / Disagreement

Participants express varying interpretations of the term "limited" in relation to \( g(x) \), and there is no consensus on the clarity of the proof or the epsilon-delta argument. The discussion remains unresolved regarding the understanding of these concepts.

Contextual Notes

There are unresolved questions about the definitions and implications of bounded versus convergent sequences, as well as the specific conditions under which the limit product rule applies.

Petrus
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Hello MHB,
Show that f(x)g(x)->0 if f(x)->0 and g(x) is limited.
The proof for this one is in the calculus book which our school use but I use 3 diffrent calculus book and can't find it in any of them and can't find it in Google, If anyone got a Link for the proof I would be glad to read it
Regards,
$$|\pi\rangle$$
 
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What do you mean by the function $g(x)$ is limited ?
 
ZaidAlyafey said:
What do you mean by the function $g(x)$ is limited ?
I mean that it's not infinity if I Also understand correct, it'S bound like $$\sin(\theta)•0$$ and $$\sin(\theta)$$ is limited so it equal 0. Does this make sense?
 
There is a difference between a sequence being bounded or convergent because a bounded sequence might not be convergent take for example $$a_n=(-1)^n$$
while the sequence is bounded $$|a_n|\leq 1$$ it doesn't converge.
 
Hello,
this is the proof that I find on My notebook and can't understand this $$|g(x)|\leq B$$
e8swar.jpg


Regards,
$$|\pi\rangle$$
 
Last edited:

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