Limit of Series fn(x): Determine Convergence Uniformly

[Bellarosa]

1. If fn (x) = (x^n)/(1+x^n) on the interval [0,∞), determine the limit from n to infinity of fn(x) and determine if it converges uniformly

Homework Equations

3. First I changed x^n = e^(n ln(x), which got no me nowhere. The pronlem that I am having is really finding the limit of an unknown costant x raisedto the variable n.

 

[Dick]

Take x=1/2 and tell me what the limit is. Now take x=2 and tell me what the limit is. Does that help? Oh, welcome to PF. I often forget to say that.

 

[Bellarosa]

the limit is 1.
Thank You.

 

[Bellarosa]

Was I right about the limit?...I was also trying to determine uniform convergence by using Weirestrass M-Test
|fn(x)|<=Mn, but Icould not find any convergent series M. I also tried finding the maximum value of the sequence fn(x) by finding it's derivative and solving for x, then substituting it into the function...Does this make sense?

 

[Dick]

Was I right about the limit?...I was also trying to determine uniform convergence by using Weirestrass M-Test
|fn(x)|<=Mn, but Icould not find any convergent series M. I also tried finding the maximum value of the sequence fn(x) by finding it's derivative and solving for x, then substituting it into the function...Does this make sense?

The limit at x=2 is 1. The limit at x=1/2 is 0. Can you generalize this to all x? Can it possibly be uniformly convergent?

 

[Bellarosa]

Ok I see the mistake I made with x = 1/2, the limit is indeed 0. I found the limit of the sequence to be 1, but based on these two values I use for x the limits are different therefore the sequence diverges...would it make sense to say that the sequence converges pointwise on [0,∞).

 

[Dick]

Ok I see the mistake I made with x = 1/2, the limit is indeed 0. I found the limit of the sequence to be 1, but based on these two values I use for x the limits are different therefore the sequence diverges...would it make sense to say that the sequence converges pointwise on [0,∞).

"the sequence diverges"?? Just because you have two different limits at two different points doesn't mean the sequence of functions diverges. Yes, it would make sense to say the sequence converges pointwise. If you can show that. What does it converge to? You can easily describe the limit function.

 

[Bellarosa]

Ok I'm a bit confused, I always thought that no matter the value it would be approaching one limit. But considering using only fractions for x will always give me a limit of 0, and if I use integers for x then then the limit is 1. Is this correct?

 

[Bellarosa]

Her is my other assumption, based on what I just wrote, if the limit of x=1/2 is 0 and the limit of x=2 is 1,fn(x) converges point wise. Because the limit ranges from [0,1] and not [0,∞) it does not not coverge uniformly. Is this logical?

 

[Dick]

Ok I'm a bit confused, I always thought that no matter the value it would be approaching one limit. But considering using only fractions for x will always give me a limit of 0, and if I use integers for x then then the limit is 1. Is this correct?

You aren't doing a very good job of describing the set of x where the limit is 0 versus the set of x where the limit is 1. pi/4 isn't a 'fraction' and pi isn't an 'integer'. But the limit at x=pi/4 is zero and the limit at x=pi is one.

 

[Dick]

Her is my other assumption, based on what I just wrote, if the limit of x=1/2 is 0 and the limit of x=2 is 1,fn(x) converges point wise. Because the limit ranges from [0,1] and not [0,∞) it does not not coverge uniformly. Is this logical?

No. That has NOTHING to do with uniform convergence. I think it's time to reread the definition.

 

[Bellarosa]

ok, Thank You.

 

[Bellarosa]

Just need a little clarification.
Is it true that a continuous function can have a discontinuous limit, but a discontinuous fuction will always have a discontinuous limit?

 

[Dick]

fn is not a 'function', it's a sequence of functions. If you actually worked this out then you should know a sequence of continuous functions can have a discontinuous limit. For the opposite case consider f(x)=1 for x>=1 and f(x)=0 for x<0. Now take fn(x)=f(x)/n. fn is pretty discontinuous, right? The limit of fn is 0. That's pretty continuous, also right?

 

[Bellarosa]

In the definition of uniform convergence particularly |fn(x) - f(x)|< epsilon, fn(x) is the sequence of funtions where as the f(x) is the function itself. Is this correct?

 

[Dick]

f is the limit of fn. If you wish to call that 'the function itself' that's your business. But I wouldn't think of it that way. In your initial example you have a sequence of nice continuous functions that converge to a step function. The limit function has different properties from the functions that are converging to it.

 

[Bellarosa]

Thank you for clarifying ...I was not thinking of it in that way it was just a bit confusing, because I read this example on line I couldn't understand why their replaced f(x) with the limit of the sequence of function and bases on that my interpretation was that f(x) was the limit of the fuction itself...