As $x$ approaches 2, $f(x)$ approaches 5. Yes, it is possible for this to be true and for $f(2)=3$.
Work Step by Step
It means that $f(x)$ becomes arbitrarily close to $3$ as we take $x$ close enough to $2$, but $x\ne2$. It is possible for $f(2)$ to equal $3$ because the limit only relies on when $f(x)$ is near $2$; there can be a hole at $x=2$.