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 ?