Proof that $a_{n}:=\sup\{|\frac{nx}{1+n^2x^2}|:x\in \mathbb{R}|\}$ $(n\in \mathbb{N})$ does not converge to $0$ I am trying to prove that the sequence $a_{n}:=\sup\big\{|\frac{nx}{1+n^2x^2}|:x\in \mathbb{R}|\big\}$ $(n\in \mathbb{N})$ does not converge to $0$. Is the following correct?
Consider arbitrary $n\in \mathbb{N}$. Choose some $x>n$. Then $$|\frac{nx}{1+n^2x^2}|\geq |\frac{n^2}{1+n^2x^2}|\geq |\frac{n^2}{n^2+n^2x^2}|=|\frac{1}{1+x^2}|:=b>0$$
Thus, for any $n\in \mathbb{N}$, $\sup\big\{|\frac{nx}{1+n^2x^2}|:x\in \mathbb{R}|\big\}\geq b>0$ (as, otherwise, for any $n \in \mathbb{N}$ we could find an $x\in \mathbb{R}$ such that $|\frac{nx}{1+n^2x^2}|>\sup\big\{|\frac{nx}{1+n^2x^2}|:x\in \mathbb{R}|\big\}$—a contradiction. Thus, $\lim_{n\rightarrow\infty} \sup\big\{|\frac{nx}{1+n^2x^2}|:x\in \mathbb{R}|\big\}>0$.
 A: As Bruno B suggested in the comments, it seems that $b$ depends on $n$. So the inequality that we would get is $a_n\geq b_n$, which is not sufficient as $b_n$ may tends to zero.
Also, since you chose $x_n$ such that $x_n>n$, we would get $b_n=\left|\frac{1}{1+x_n^2}\right|\leq \left|\frac{1}{1+n^2}\right|\xrightarrow{n\to\infty} 0$. So choosing $x$ such that $x>n$ would not work in this case.
Note that
$$
a_n=\sup\left\{\left|\frac{nx}{1+n^2x^2}\right|:x\in \mathbb{R}\right\}\geq \frac{n\cdot \frac{1}{n}}{1+n^2\cdot \frac{1}{n^2}}=\frac{1}{2}
$$
Thus, because limit preserve weak inequality, if $a_n$ converges, it will uphold
$$
\lim_{n\to\infty} a_n\geq \frac{1}{2}
$$
A: You can assume $x\ge0$. Let
$$ f_n(x)=\frac{nx}{1+n^2x^2}. $$
Then
$$ f_n'(x)=\frac{n(1-n^2x^2)}{(1+n^2x^2)^2}. $$
Letting $f_n'(x)=0$ gives $x=:x_n\equiv\frac1n$. Clearly
$$ f_n''(x_n)=-\frac{n^2}{2}<0 $$
and hence $f_n(x)$ attains a local maximum at $x=x_n$. Since $f_n(0)=f_n(\infty)=0$, you can conclude that $f_n(x)$ attains the global maximum at $x=x_n$. Therefore
$$ a_n=\sup\big\{|\frac{nx}{1+n^2x^2}|:x\in \mathbb{R}\big\}=f_n(x_n)=\frac12 $$
which implies $a_n$ do not converge to $0$.
