Prove whether or not the function [math] f(x)=\frac{1}{x+1}[/math] is uniformly continuous in [math](0,\infty)[/math]

The only discontinuity is at x=-1, which is not included in the domain (0,infinity). Therefore the function is uniformly continuous in(0,infinity).

The only discontinuity is at x=-1, which is not included in the domain (0,infinity). Therefore the function is uniformly continuous in(0,infinity).

 

Your conclusion is based on what definition or theorem?

That a function to be continuous on an interval, it has to be continuous at every point in the interval.

That a function to be continuous on an interval, it has to be continuous at every point in the interval.

..

 

That is the definition of a function being continuous on an interval and not the definition of a function being uniformly continuous over an interval

Sorry, another factor is remaining which is lim f(x), as x tends to infinity, exists and is equal to zero.

 

And, we have the theorm "If a real-valued function f(x) is continuous on (0,infinity) and its limit, as x tends to infinity, exists (and is finite), then f(x) is uniformly continuous."

Sorry, another factor is remaining which is lim f(x), as x tends to infinity, exists and is equal to zero.

 

And, we have the theorm "If a real-valued function f(x) is continuous on (0,infinity) and its limit, as x tends to infinity, exists (and is finite), then f(x) is uniformly continuous."

 

It seams to me that you do not know the definition of uniform continuity.

 

As for the theorem you stated it exists nowhere in the whole of cosmos.

 

 

Anyway if you still insist on the validity of your theorem ,then give a proof.

Check the following: http://en.wikipedia.org/wiki/Uniform%5Fcontinuity

Go to the last point in "Properties" section.

Check the following: http://en.wikipedia.org/wiki/Uniform%5Fcontinuity

Go to the last point in "Properties" section.

.

 

I see you have a problem in reading :

 

The section under properties it says that:

 

if the interval on which A continuous function f is defined is COMPACT then the function is uniformly continuous.

 

Is the interval (0,[math]\infty[/math]) on which f is defined COMPACT???

"Compact interval" is not involved here. It is a separate point.

"Compact interval" is not involved here. It is a separate point.

 

Yes you can escape by writing nonsenses because the moderators of this forum

 

perhaps are not capable of detecting such unfounded mathematical nonsenses that you keep on writing.

Thanks.

 

And, we have the theorm "If a real-valued function f(x) is continuous on (0,infinity) and its limit, as x tends to infinity, exists (and is finite), then f(x) is uniformly continuous."

 

Can you find anywhere such a theorem?

Yes you can escape by writing nonsenses because the moderators of this forum

 

perhaps are not capable of detecting such unfounded mathematical nonsenses that you keep on writing.

Triclino, you can make your argument and your counter-argument without being rude. Otherwise, spare everyone and don't make an argument at all.

 

This attitude is unacceptable.

Yes you can escape by writing nonsenses because the moderators of this forum

 

perhaps are not capable of detecting such unfounded mathematical nonsenses that you keep on writing.

 

triclino, there are several members of the board (myself included) that know perfectly well what uniform continuity is - and indeed pretty much everything else you've posted about so far. I covered it myself about 6 years ago.

 

I don't know whether you genuinely cannot answer your problem or simply wish to massage your ego by posting problems you already know the answer to. It doesn't really matter in any case - you need to make careful note of two things if you wish to remain a member of these boards:

 

• We are not here to answer your homework problems. If you have a proof already which you want checking, then post it.
  
• If you disagree with another member's point, then do so in a courteous fashion. As mooey has pointed out, it's totally unnecessary.