Uniform convergence of sequence of functions

phosgene
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Homework Statement



Let [itex]f_{n}(x)=\frac{x}{1+x^n}[/itex] for [itex]x \in [0,∞)[/itex] and [itex]n \in N[/itex]. Find the pointwise limit f of this sequence on the given interval and show that [itex](f_{n})[/itex] 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 :

[itex]0[/itex] if [itex]x=0[/itex]
[itex]x[/itex] if [itex]0<x<1[/itex]
[itex]1/2[/itex] if [itex]x=1[/itex]
[itex]0[/itex] if [itex]x>1[/itex]

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 [itex]|f_{n}(x) - f(x)| ≥ ε[/itex], but I'm not sure how to go about this.
 
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phosgene said:

Homework Statement



Let [itex]f_{n}(x)=\frac{x}{1+x^n}[/itex] for [itex]x \in [0,∞)[/itex] and [itex]n \in N[/itex]. Find the pointwise limit f of this sequence on the given interval and show that [itex](f_{n})[/itex] 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 :

[itex]0[/itex] if [itex]x=0[/itex]
[itex]x[/itex] if [itex]0<x<1[/itex]
[itex]1/2[/itex] if [itex]x=1[/itex]
[itex]0[/itex] if [itex]x>1[/itex]

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 [itex]|f_{n}(x) - f(x)| ≥ ε[/itex], but I'm not sure how to go about this.

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

Now choose x = e +1?
 

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