Solution: At first we claim that the sequence $\left\{ f_n(x) \right\} $ converges to $f(x)=0,~\forall x\in \mathbb{R} $ pointwise. Recall that,

Here $D$ is the domain of function. In the above definition, note that the choice of $n_0$ depends upon both $x$ and $\varepsilon $. In summary, the sequence $\left\{ f_n(x) \right\} $ converges to $f(x)$ pointwise means for a fixed $x\in D$ the sequence of numbers $\left\{ f_n(x) \right\} $ should converge to $f(x)$.

In order to prove the given sequence of functions converges to the zero function pointwise, let $\varepsilon >0$ be given and $x\in \mathbb{R} \setminus \{0\}$. It is clear that $x=0$, the sequence $f_n(x)=0$ and this converges to $0$ being a constant sequence. Note that \begin{align*} 1+n^2 x^2 > n^2 x^2 & \implies \frac{1}{1+n^2 x^2} \lt \frac{1}{n^2 x^{2} } \\ & \implies \frac{nx}{1+n^2 x^{2} } \lt \frac{nx}{n^{2} x^{2} } = \frac{1}{nx}. \end{align*} By the Archimedean property, we choose $n_0$ such that $n_0>\frac{1}{x\varepsilon }$. Thus, for $n\geq n_0$ we have \begin{align*} \left\vert f_n(x) - f(x) \right\vert & = \left\vert \frac{nx}{1+n^2 x^2} \right\vert \\ & \lt \left\vert \frac{1}{nx} \right\vert \\ & \leq \left\vert \frac{1}{n_0 x} \right\vert \\ & \lt \varepsilon . \end{align*} Thus, $\left\{ f_n(x) \right\} $ converges to $f(x)$ pointwise.

We now claim that the convergence is not uniform. We recall that

Let us suppose that the sequence converges to $f(x)$ uniformly. That means for any $\varepsilon >0$, \eqref{eq:12Mar2023-1} should hold. Take $\varepsilon =\frac{1}{2}$. For this $\varepsilon $ we can find $n_0\in \mathbb{N} $ such that for all $x\in \mathbb{R} $ we have \[ \left\vert \frac{n_0x}{1+n_0^2 x^{2} } \right\vert < \frac{1}{2} . \] Since the above holds for all $x$, we take $x=\frac{1}{n_0}$. Note that \[ \left\vert \frac{n_0x}{1+n_0^2 x^2} \right\vert = \left\vert \frac{1}{1+1} \right\vert = \frac{1}{2} < \frac{1}{2} \] is absurd and hence the convergence can not be uniform.