I am trying to disprove uniform convergency but without any success, and I dont know how to approach it.
I am trying to prove that the following function : $$f_n(x)=\frac{nx}{1+(nx)^2}$$ is not uniformly convergent in $[0,1]$ by defintion that is to say:
$$\exists \epsilon : \forall N \in \mathbb{N} : \exists n \geq N: \exists x \in [0,1] : |f_n(x)-f(x)| \geq \epsilon $$
My efforts so far:
first I found $f(x)$: $$f(x)=\lim_{n \to \infty} \frac{nx}{1+(nx)^2}= \Biggl\{\begin{array}{l}\frac{0}{1+0}=0, x=0\\0,\ \ \ \ \ \ \ \ \ \ \ ,x \neq 0 \end{array} = 0 $$
Threfore I need to work with:
$$|f_n(x)-f(x)| = \left| \frac{nx}{1+(nx)^2} - 0\right| = \left| \frac{nx}{1+(nx)^2} \right|$$
Consider $\epsilon=\frac{1}{2}$ and for any $N\geq 1$ we have $n=N$ and $x=\frac{1}{N}\in[0,1]$ we have $|f_n(x)|=\frac{1}{2}$.