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I've been asked to prove that the sequence $[f_n]_{n \in N}$ with $f_{n}(x)=\frac{x^{2n}}{1+x^{2n}}$ converges uniformly on $x \in [1+\delta,\infty[$ where $\delta > 0 $.

So far I've found that $f(x)=0$ for $x \in ]-1,1[$, $f(x)=\frac{1}{2}$ for $x=\pm 1$ and $f(x)=1$ for $x \notin [-1,1]$. I have calculated the distance between each of them and found it to be $\frac{1}{2}$. It was 0 for $x=\pm 1$. I am told, that I need to prove it using some sort of epsilon/delta proof, however I've been unsuccesful so far. Any help would be greatly appreciated.

Bérénice
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O. T.
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  • I suspect you also have a problem in understanding nested quantifiers. See http://math.stackexchange.com/a/813372/21820 for an explanation of uniform convergence and logical structure. – user21820 May 10 '16 at 13:55

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Hint

Note that $\frac{x^{2n}}{1+x^{2n}} = 1 - \frac{1}{1+x^{2n}} \in 1-\left[0,\frac{1}{1+(1+δ)^{2n}}\right]$ for any $x \in [1+δ,\infty)$.

So it suffices to show that $\frac{1}{1+(1+δ)^{2n}} \to 0$ as $n \to \infty$.

user21820
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  • Perfect. That would be the same as proving that a delta exists such that $\varepsilon >\frac{1}{(1+\delta)^{2n}}$. Correct? – O. T. May 10 '16 at 12:18
  • @O.T.: Not sure what you mean. Uniform convergence does not have any delta; it's that for any error margin, from a certain point onwards in the sequence of functions it is close to the pointwise limit (to within the error margin). – user21820 May 10 '16 at 13:03
  • @O.T.: I think I get your question. No of course not. $δ$ is given in your question right at the start; you can't change it. I said "as $n \to \infty$", which is exactly what you want (see my previous comment). You should learn to think in terms of asymptotic behaviour, not just formal definitions; anyway the last line in my answer says "when $n$ is big enough, $\frac{1}{1+(1+δ)^{2n}}$ would be small enough". No matter how small you want the latter, when $n$ is big enough you will get what you want. – user21820 May 10 '16 at 13:50
  • Yea, that was my question. Eventually I figured, that I could just isolate N in this equation $\frac{1}{1+(1+\delta)^{2N}}<\varepsilon$. – O. T. May 14 '16 at 09:33
  • @O.T.: Correct. $δ$ is a constant so you can find right at the beginning some natural $N$ such that $\frac{1}{1+(1+δ)^{2n}}<ε$ for any natural $n \ge N$. – user21820 May 14 '16 at 09:42