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 ?

Think of it this way

\(\displaystyle \lim_{ t \to 0} \) \(\displaystyle tsin ( \frac{1}{t} ) \)

Let us approach 0 from the posative side

\(\displaystyle \lim_{ t \to 0^{+}} tsin ( \frac{1}{t} ) \)

This will make the limit

\(\displaystyle \lim_{ t \to 0^{+}} (0^{+} )sin ( \infty ) \)

Note that \(\displaystyle sint \) is a sinoncidal function and \(\displaystyle sin \infty \) will evaluate to some number from 0 to 1.

However, even if sin WASNT BOUNDED by 0 and 1...say \(\displaystyle sin \infty = n \) where \(\displaystyle n>1 \)

When we multiply by 0, i.e.

\(\displaystyle 0 * n \)

It is equal to 0!

\(\displaystyle 0 * n = 0\)

So you can see the above limit

\(\displaystyle \lim_{ t \to 0^{+}} tsin ( \frac{1}{t} ) = 0 \)