You need to use squeeze theorem to answer this questions. Breaking the function into parts, the lim as x → 0 of 2pi/x is infinity, so we have sine function going to infinity which is alternating between -1 and +1 or -1 < sin(2pi/x) < 1 now the lim x → 0 of x^3 is 0 so when you multiply 0 times the +/- 1 you get 0.
Therefore lim x→0 x^3 sin(2pi/x) = 0