\b[False.] The bounded continuous function $f(x)=\sin(1/x)$ on $(0,1]$ has no limit at zero. If polynomials converged uniformly to it there, they would be uniformly Cauchy on $(0,1]$ and hence on $[0,1]$ by continuity at zero. Their uniform limit on $[0,1]$ would continuously extend $f$ to zero, which is impossible.


Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 11F / b / iii / Solution

 Solution (source code)