Let $f_n(x)=nx^n$ , $x \in [0,1)$.
It is given that ${f_n}$ is pointwise convergent. It is clear that $f_n(0)=0$, and $f_n(x)=0$ ,if $x\in (0,1)$. But even after excluding $1$ from the domain, points near $1$, the function may exhibit bad behavior. For example:
If $x_n= \sqrt[n]{n} \to 1$ and then $f(x_n)=n^2$, which tends to $\infty$. Again if we take $x_n=\frac{1}{\sqrt[n]{n}}$ then $f(x_n)=1$.
As $f(x_n)$ may tend to $\infty$, how can we say that ${f_n}$ is pointwise convergent?
Pointwise convergence means that if you fix $x_0 \in [0, 1)$, then $f_n(x_0) \rightarrow f(x_0)$ where $f$ is the pointwise limit of $f_n$. In your case, if you fix $x \in [0, 1)$, you know that $n x^n \rightarrow 0$, when $n$ goes to $+\infty$ (and $x$ is fixed, it is point wise convergence). Therefore you can say that $(f_n)$ converges pointwise to $0$.
When you consider pointwise convergence you cannot choose $x$ as a function of $n$ because pointwise convergence precisely means that you look at the convergence for a fixed $x$.
However $(f_n)$ does not converge uniformly to $0$ on $[0, 1)$, and you can see it for example by looking at $f_n(1-1/n)=n(1-1/n)^n \sim n/e \rightarrow + \infty$. Nevertheless uniform convergence is a stronger notion than pointwise convergence.