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I am having trouble trying to prove that the sequence of functions (fn) defined on the interval [0,1] by fn(x)=nx/(1+(nx^2)) does not converge uniformly. Namely, my difficulty is coming from choosing an epsilon.

Thanks for any help in advance

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  • $\begingroup$ Choose $\varepsilon = \frac12$. It's a nice number and works. $\endgroup$ – Daniel Fischer Jan 11 '14 at 21:35
  • $\begingroup$ Any $\epsilon$ will work. For $x>0$, $\bigl|{nx\over 1+nx^2}-{1\over x}\bigr|= {1\over x (1+nx^2) }$. Consider the limit of this expression as $x\rightarrow0^+$. $\endgroup$ – David Mitra Jan 11 '14 at 21:43
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Depending on what theorems you have at your disposal, there's another way to go about this that does not require getting your hands dirty.

If a sequence $(f_n)$ of functions converges uniformly to $f$, then $f$ is also continuous.

In this particular case, a quick check shows that $$ \lim_{n\rightarrow\infty}f_n(x)=\begin{cases}\frac{1}{x} & \text{if }x\neq 0\\ 0 & \text{if }x=0\end{cases}, $$ which is not continuous. So, if you can (formally) show that pointwise limit, then you should be all set.

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The limit of the sequence $(f_n)$ is the function $f$ defined by $$f(0)=0\quad\text{and}\quad f(x)=\frac 1 x,\; x\ne 0$$ so since $f$ isn't continuous on $[0,1]$ while $f_n$ are continuous so the convergence isn't uniform.

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