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Due to periodicity in $x$ you can write $f$ as a function on $(0,\infty)\times S^1$ so that your assumption in 1. becomes just uniform convergence on $S^1$ and this implies $f(t/\varepsilon,\cdot)\to a$ uniformly and hence $f(t/\varepsilon,x/\epsilon)\to a$. The same argument gives the second statement.