A series of functions

Pointwise. For $x=0$, all terms vanish. For $x>0$, $1+nx\ge nx$, so
$$0\le u_n(x)=\frac1n\cdot\frac x{1+nx}\le\frac1n\cdot\frac1n=\frac1{n^2}.$$
The series converges.

Normally. The bound $0\le u_n(x)\le\frac1{n^2}$ holds for all $x\ge0$, and $\sum\frac1{n^2}<\infty$. So $\sum\sup_x|u_n|<\infty$: the convergence is normal, hence uniform, on $[0,+\infty)$.

Consequence. Each $u_n$ is continuous on $[0,+\infty)$, and a uniformly convergent series of continuous functions has a continuous sum. So $f$ is continuous on $[0,+\infty)$, with $f(0)=0$.