Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 1 / 4G / ii / Solution

For every fixed $x>0$, $n{x}^2\to \infty$, so the pointwise limit is zero. But ${\sup }_{0<x<1}{e}^{-n{x}^2}=1$ for every $n$, since $x$ can approach zero. Hence convergence is not uniform. For an explicit moving-point test take $x_n=n^{-1/2}$ for $n>1$, which gives $F_n(x_n)=e^{-1}$.