To prove this, simply do the following:
Sin[1/t] t^2 = Sin[1/t]/(1/t) t
Take the limit:
Limit[Sin[1/t]/(1/t) t, t -> 0] = Limit[Sin[1/t]/(1/t), t -> 0] Limit[t, t -> 0] = 1*0
And you'll get your answer: 0
Also, interesting implications about the series of a bunch of reciprocals converging to a single point as they all approach the same pole.
