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Show that the sequence $\{f_n\}$ of function where $f_n(x)=n/(x+n)$, is uniformly convergent in $[0,k]$ whatever $k$ may be, but not uniformly convergent in $[0,\infty)$.

The sequence is point wise convergent $\forall x\geq 0$, $$f(x)=1\hspace{1 cm} \forall x\geq0$$

How to proceed further?

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    $\begingroup$ Start with calculating $f(x) - \frac{n}{n+x}$. Look at the result. $\endgroup$ Oct 19 '17 at 19:18
  • $\begingroup$ Actually, $\lim\limits_{x\to +\infty} \frac{x}{x+n} = 1$. What is the maximal value that $\frac{x}{x+n}$ attains on $[0,k]$? $\endgroup$ Oct 19 '17 at 19:37
  • $\begingroup$ @DanielFischer $k/(k+n)$ $\endgroup$
    – Vincent
    Oct 19 '17 at 19:45
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    $\begingroup$ No, you're confusing the roles of $k$ and $n$. Maybe it would have been better to us $a$ instead of $k$. You have $$0 \leqslant 1 - \frac{n}{x+n} \leqslant \frac{k}{k+n}$$ on $[0,k]$. Given $\varepsilon > 0$, you want $\frac{k}{k+n} \leqslant \varepsilon$ for all large enough $n$. $\endgroup$ Oct 19 '17 at 20:07
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The pointwise limit is $f(x)=1$.

In $[0,k]$ we have $$|f(x)-f_n(x)|=\left|1-\frac{n}{x+n}\right|=\frac{x}{x+n}$$

The map $$x\mapsto\frac{x}{x+n}$$ is an increasing function on $[0,k]$. So the maxima of this function (which occurs at $x=k$) is $\frac{k}{k+n}$.

So $$\lim_{n\to\infty}(\sup_{x\in[0,k]}|f(x)-f_n(x)|)= \lim_{n\to\infty}\frac{k}{k+n}=0$$ Hence $f_n\rightarrow f$ uniformly in $[0,k]$.

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