Uniform convergence of sequence of functions

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SUMMARY

The discussion focuses on the sequence of functions defined by f_{n}(x)=\frac{x}{1+x^n} for x in [0,∞) and n in N. The pointwise limit of this sequence is established as follows: f(0)=0, f(x)=x for 01. The participants highlight the challenge of proving that the sequence does not converge uniformly to the limit function, specifically by demonstrating that for any natural number N, there exists an x such that |f_{n}(x) - f(x)| ≥ ε.

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phosgene
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Homework Statement



Let f_{n}(x)=\frac{x}{1+x^n} for x \in [0,∞) and n \in N. Find the pointwise limit f of this sequence on the given interval and show that (f_{n}) does not uniformly converge to f on the given interval.

Homework Equations


The Attempt at a Solution



I found that the pointwise limit f is :

0 if x=0
x if 0<x<1
1/2 if x=1
0 if x>1

But I'm stuck on proving that it's not uniformly convergent. I know that for any natural number N, I need to find some x such that |f_{n}(x) - f(x)| ≥ ε, but I'm not sure how to go about this.
 
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phosgene said:

Homework Statement



Let f_{n}(x)=\frac{x}{1+x^n} for x \in [0,∞) and n \in N. Find the pointwise limit f of this sequence on the given interval and show that (f_{n}) does not uniformly converge to f on the given interval.

Homework Equations





The Attempt at a Solution



I found that the pointwise limit f is :

0 if x=0
x if 0<x<1
1/2 if x=1
0 if x>1

But I'm stuck on proving that it's not uniformly convergent. I know that for any natural number N, I need to find some x such that |f_{n}(x) - f(x)| ≥ ε, but I'm not sure how to go about this.

Look at |f_n(x) - f(x)| for x close to, but not equal to, 1.
 
\frac{x}{1+x^n}-x = \frac{x^{n+1} }{1+x^n} \geq \frac{x^{N+1} }{1+x^N} \geq x

Now choose x = e +1?
 

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