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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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    A [related problem](http://math.stackexchange.com/questions/370023/how-to-prove-a-sequence-of-a-function-converges-uniformly/370071#370071).2013-06-10

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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 $\mathbb R$.