This is probably a very silly question. But I hope you'll help me with it.

Lim (t -> 0) tsin(1/t) = 0

I understood why it is 0, because of the squeeze principle.

But sin(1/t) can be expanded as (1/t) - (1/3!t^3) + ...

So tsin(1/t) = 1 ignoring higher powers right as t->0.

Or I cant ignore the hgher powers because as t -> 0, 1/t approaches infinity.

Am I making sense ?