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Show that $$f_n: \mathbb{R} \rightarrow \mathbb{R}; \ n\ge1; \ f_n (x)=\frac{x \sqrt{n}}{n \sqrt{n}+x^2}$$ does not converge uniformly on $\mathbb{R}$

I have showed that $f_n(x)$ pointwise converges to $0$.

Then I find the maximum of $|f_n(x)|$ and it is for $x=n^{3/4}$ and $x=-n^{3/4}$

Hence $\sup_{\mathbb{R}}{|f_n(x)|}=f_n(n^{3/4})=\dfrac{1}{2 n^{1/4}} \rightarrow 0$

Where is the mistake?

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There is no mistake, the convergence is uniform! – Mercy King Dec 6 '12 at 20:16
then is wrong the text of the exercise.... – Madara Dec 6 '12 at 20:31
A related problem. – Mhenni Benghorbal Jun 10 '13 at 9:27

This post is made community wiki in order to remove this question from the "unanswered" list

If what you say is true, the text is wrong. You have shown that the sequence of functions does indeed converge uniformly on $\mathbb R$.

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