elementary Uniform convergence question $f_{n}(x)=\{\frac{nx}{1+n^2x^2}\}$ , $0\le x\le 1$
So to find the limit function (If it called this way) we take $x_{0}$ and the limit, so we get $\lim_{n\to \infty}f_n(x)=\lim_{n\to\infty}\frac{nx_{0}}{1+n^2x_{0}^2}=\lim_{n\to \infty}\frac{\frac{nx_{0}}{n^2}}{\frac{1}{n^2}+\frac{n^2x_{0}^2}{n^2}}=\lim_{n\to \infty}\frac{\frac{x_0}{n}}{\frac{1}{n^2}+x_{0}^2}=0$
Nowe to find Uniform convergence we let there be  $\epsilon>0$ we need to find $N<n$ such that $|f_n(x)-f(x)|<\epsilon$
So we have $|\frac{nx}{1+n^2x^2}-0|=\frac{nx}{1+n^2x^2}<\frac{nx}{n^2x^2}=\frac{1}{nx}<\epsilon$
So we got $\epsilon$ that depends on $n$ and $x$ so therefore it is not uniform converges.
The book gives an example of the point $\frac{1}{n}$, what can we learn from this point, or what does this point proves?   
 A: The book gives an example of the point $\frac{1}{n}$, what can we learn from this point, or what does this point proves? 
One may recall that uniform convergence is defined in terms of $\sup_{I} \left|f_n(x)-f(x) \right|$.
One may observe that,
$$
f_n'(x)=\frac{n(1+nx)(1-nx)}{\left(1+n^2 x^2\right)^2}
$$  giving that, $f$ reaches its maximum for $x=\frac1n$, then
$$
\sup_{x \in [0,1]} \left|f_n(x) -0\right|=\left|f_n \left(\frac1n \right) \right|=\frac12 \neq0.
$$
A: 'So we got ϵ that depends on n and x so therefore it is not uniform converges.'
You can't conclude that like that. 
One way to show that convergence is not uniform is to show that $$\lim_{n\to\infty}\sup_{x\in [0,1]}|f_n(x)-f(x)|\neq 0.$$
It is easy to check that $$\sup_{x\in [0,1]}|f_n(x)-f(x)|=\max_{x\in [0,1]}{|f_n(x)|}=|f_n(\frac{1}{n})|=\frac{1}{2}.$$ 
(Max is attained in end-points $0$, $1$ or in point where $f'_n(x)=0$. By checking these cases we find $\max_{x\in [0,1]}|f_n(x)|=\frac{1}{2}$ is attained in point $\frac{1}{n}$.)
Therefore,
$$\lim_{n\to\infty}\sup_{x\in [0,1]}|f_n(x)-f(x)|=\frac{1}{2}\neq 0$$
and we can conclude that convergence isn't uniform.
A: To prove that a sequence of functions $(f_n)$ does not converge uniformly to $0$, you can not just write a sequence of inequalities and affirm that at the end it depends on $x$. What proves that you could have an $\epsilon$ not depending on $x$ at the end if you did some other sequence of inequalities ?
You have two ways of tackling the problem :


*

*You can study the function $f_n$ and get its maximum : it is reached for $x=1/n$. Thus, $\sup_{x\in[0,1]} |f_n(x)| = f_n(1/n) = 1/2$ and it does not converge uniformly to $0$.

*If $(f_n)$ converges uniformly to $f$, then for every sequence $(x_n)$ converging to $x$, $(f_n(x_n))$ converges to $f(x)$. With $x_n=1/n$, $f_n(x_n) = 1/2 \rightarrow 1/2$ and $f(1/2)=0$. So $(f_n)$ does not converge uniformly to $0$.

