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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 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
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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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