Uniform convergence of $f_n(x)=\frac{n\cdot x}{1+n^2}$ I want to check the uniform convergence of $\displaystyle{f_n(x)=\frac{n\cdot x}{1+n^2}}$.
We have that \begin{equation*}f^{\star}=\lim_{n\rightarrow \infty}f_n(x)=\lim_{n\rightarrow \infty}\frac{n\cdot x}{1+n^2}=\lim_{n\rightarrow \infty}\frac{n\cdot x}{n\left (
\frac{1}{n}+n\right )}=\lim_{n\rightarrow \infty}\frac{x}{
\frac{1}{n}+n}=0\end{equation*}
So $f_n$ converges $f^{\star}=0$.
The supremum of $|f_n(x)-f^{\star}|$ on $\mathbb{R}$  is infinity and so we don't have uniform convergence, right?

Recall the definition of uniform convergence:

A sequence of functions $\{f_n\}$ on a set $E$ is said to converge uniformly to $f$ if for all $\epsilon>0 \exists N_0$ s.t. $$|f_n(x)-f(x)|<\epsilon$$ for all $n>N_0$ and for all $x\in E$.

Now, in this case, $$|f_n(x)-f(x)|<\epsilon$$
$$\implies \frac{n|x|}{1+n^2}<\epsilon$$
Lets, fix $\epsilon=1/2$ and $E=\mathbb R^+$. So, this translates to $$\frac{nx}{1+n^2}<nx<\frac{1}{2}$$ What about $n>\left \lfloor\frac{1}{2x}\right \rfloor $? Is selection of $N_0$ independent of selection of $x$ as required by the definition?
