If I have that $f_n(x) = \frac{nx}{1+nx^3} , n=1,2...$ and $f(x)=\lim_{n \to \infty}f_n(x)$ and the question is: "determine $f$", does that mean that I have to compute the limit?, i.e. to compute $f(x)=\lim_{n \to \infty}f_n(x)$ ?
And also how should I think about the convergence uniform on an interval?
I'm assuming you're working on $[0,+\infty\rangle$ to avoid zeroes in the denominator.
Pointwise for $x \in [0,+\infty\rangle$ we have $$\lim_{n\to\infty} f_n(x)=\lim_{n\to\infty} \frac{nx}{1+nx^3} = \begin{cases} \frac1{x^2}, &\text{ if } x > 0\\ 0, &\text{ if }x=0\end{cases}$$
so $f(x) = \frac1{x^2} \chi_{\langle 0,+\infty\rangle}(x)$.
On an interval $[a,b] \subseteq \langle 0, +\infty\rangle$ the convergence is uniform. This follows from the fact that $(f_n)_n$ is monotonically increasing to $f$ (i.e. $f_n(x) \le f_{n+1}(x) \le f(x)$ for all $n \in \Bbb{N}$ and $x \in [a,b]$) and $f$ is continuous on $[a,b]$. Uniform convergence follows from Dini's theorem.
On an interval of the form $[0,b]$ we cannot make such a conclusion since $f$ is not continuous. Indeed, the convergence on $[0,b]$ is not uniform since for large enough $n\in\Bbb{N}$ we have $\frac1n \in [0,b]$ and $$\left|f\left(\frac1n\right)-f_n\left(\frac1n\right)\right| = n^2 - \frac1{1+\frac1{n^2}} \xrightarrow{n\to\infty} +\infty.$$