1. ## Derivative

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 ?

2. Originally Posted by poorna
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

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

Let us approach 0 from the posative side

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

This will make the limit

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

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

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

When we multiply by 0, i.e.

$0 * n$

It is equal to 0!

$0 * n = 0$

So you can see the above limit

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

3. Originally Posted by poorna
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 ?
Yes, you can't "ignore higher powers" for the precise reason you say- they do not become smaller as t goes to 0.