Convergence of $f_n(x) = \frac{nx}{1+n^2x^2}$ (Disclaimer: This is a homework question.)
$$\text{For }n\text{ in }\mathbb{N}\text{ let }f_{n}:\mathbb{R} \rightarrow \mathbb{R} \text{ be defined by } f_{n}(x)=\frac{nx}{1+n^2x^2}
$$

Show that the sequence $(f_{n})_{n\in\mathbb{N}}$ is convergent. Is this convergence uniform?

My attempt:
as $n \rightarrow \infty\quad f_{n} \rightarrow \frac{1}{nx}  \implies f_{n} \rightarrow f = \left\{\begin{matrix}
 &0  &x\neq0 \\ 
 &??  &x = 0 
\end{matrix}\right.
$
It seems to me $f$ is discontinous at $x=0$, unless I misunderstood.
For uniform convergence it's required that:$$\forall \varepsilon>0\quad \exists K_{\varepsilon}\in\mathbb{N} \quad s.t.\quad \forall n\geq K_{\varepsilon}\\||f_{n}(x)-f(x)||<\varepsilon$$
if we pick $n=k$ and $x_{k} = \frac{1}{k}$ then $$f_{n} = 1\quad \forall n\in\mathbb{N}\\\implies||f_{n}(x)-f(x)|| = ||1-0||=1$$
Does this imply $f_{n}$ is not uniformly convergent?  
 A: Recall that a sequence $(f_n)_{n=1}^\infty$ converges pointwise to $f$ on $\mathbb{R}$ if for every $x\in \mathbb{R},$ the sequence of scalars $(f_n(x))_{n=1}^\infty$ converges to $f(x).$
Also, the sequence $(f_n)_{n=1}^\infty$ converges uniformly to $f$ on $\mathbb{R}$ if for every $\varepsilon>0,$ there exists a natural number $N_\varepsilon$ such that for any $n\geq N_\varepsilon$ and any $x\in \mathbb{R},$ we have 
$$| f_n(x)-f(x)    |<\varepsilon.$$
If  $f_n(x) = \frac{nx}{1+n^2x^2},$ then the sequence $(f_n)_{n=1}^\infty$ converges pointwise to $f=0$ on $\mathbb{R}.$
Indeed, fix $x\in \mathbb{R}.$
Then 
$$\lim_{n\to\infty}f_n(x) = \lim_{n\rightarrow\infty}\frac{nx}{1+n^2x^2} = \lim_{n\to\infty}\frac{\frac{x}{n}}{\frac{1}{n^2}+x^2} = 0 = f(x).$$
So the sequence $(f_n)_{n=1}^\infty$ converges pointwise on $\mathbb{R}.$
Note that the sequence $(f_n)_{n=1}^\infty$ does not converge uniformly to $f=0$ on $\mathbb{R}.$
Let $\varepsilon=\frac{1}{4}.$
Fix $N_\varepsilon\in\mathbb{N}.$
Choose $n = N_\varepsilon$ and $x = \frac{1}{N}.$
Then 
$$| f_n(x) - f(x) | = \bigg| \frac{1}{1+1} - 0 \bigg| = \frac{1}{2} > \frac{1}{4} = \varepsilon.$$
So the sequence $(f_n)_{n=1}^\infty$ does not converge uniformly to $f=0$ on $\mathbb{R}.$
