Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 3 / 12F / b / i

True. Given $\epsilon >0$, uniform continuity supplies $\delta >0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$. For all sufficiently large $n$, $1/n<\delta$, and thenThis is precisely uniform convergence of $f_n$ to $f$.