Show that f(x)g(x)->0

Petrus

Well-known member
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 cant 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,
$$\displaystyle |\pi\rangle$$

Staff member

johng

Well-known member
MHB Math Helper
Hi,
The attachment gives a short proof of your problem:

ZaidAlyafey

Well-known member
MHB Math Helper
What do you mean by the function $g(x)$ is limited ?

Petrus

Well-known member
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 $$\displaystyle \sin(\theta)•0$$ and $$\displaystyle \sin(\theta)$$ is limited so it equal 0. Does this make sense?

ZaidAlyafey

Well-known member
MHB Math Helper
There is a difference between a sequence being bounded or convergent because a bounded sequence might not be convergent take for example $$\displaystyle a_n=(-1)^n$$
while the sequence is bounded $$\displaystyle |a_n|\leq 1$$ it doesn't converge.

Petrus

Well-known member
Hello,
this is the proof that I find on My notebook and cant understand this $$\displaystyle |g(x)|\leq B$$

Regards,
$$\displaystyle |\pi\rangle$$

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