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

[Petrus]

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,
$$|\pi\rangle$$

 

[MarkFL]

There is a proof of the product rule for limits given here:

Calculus/Proofs of Some Basic Limit Rules - Wikibooks, open books for an open world

 

[Petrus]

There is a proof of the product rule for limits given here:

Calculus/Proofs of Some Basic Limit Rules - Wikibooks, open books for an open world

Hmm..? That Was a confusing one but I think I am suposed to show $$f(x)g(x)<\epsilon$$ cause $$\epsilon>0$$ right? In the delta epsilon proof we say $$\epsilon>0$$ I am wrong or?

Regards,
$$|\pi\rangle$$

 

[johng1]

Hi,
The attachment gives a short proof of your problem:

View attachment 1348

 

[alyafey22]

What do you mean by the function $g(x)$ is limited ?

 

[Petrus]

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?

 

[alyafey22]

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.

 

[Petrus]

Hello,
this is the proof that I find on My notebook and can't understand this $$|g(x)|\leq B$$

Regards,
$$|\pi\rangle$$