Welcome to our community

Be a part of something great, join today!

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

Petrus

Well-known member
Feb 21, 2013
739
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,
\(\displaystyle |\pi\rangle\)
 

Petrus

Well-known member
Feb 21, 2013
739

johng

Well-known member
MHB Math Helper
Jan 25, 2013
236
Hi,
The attachment gives a short proof of your problem:

MHBcalculus1.png
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
What do you mean by the function $g(x)$ is limited ?
 

Petrus

Well-known member
Feb 21, 2013
739
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
Jan 17, 2013
1,667
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
Feb 21, 2013
739
Hello,
this is the proof that I find on My notebook and cant understand this \(\displaystyle |g(x)|\leq B\)


Regards,
\(\displaystyle |\pi\rangle\)
 
Last edited: