Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 1 / 1E / a / Solution

No. Consider the triangular functions $f_n(x)=\max (1-n|x|,0)$. The absolute-value and positive-part maps are Lipschitz continuous, so $|f_n(x)-f_n(y)|\le n|x-y|$. Each $f_n$ is therefore uniformly continuous on the whole real line. For $x=0$ its value is always one; for every fixed $x\ne 0$ it eventually vanishes. Hence its pointwise limit isThis limit is not continuous at zero: along $x_j=1/j\to 0$, $f(x_j)=0\nrightarrow 1=f(0)$. Individual uniform continuity does not provide a modulus uniform in $n$.