# Derivative

#### 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 ?

#### AllanCuz

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

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

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• poorna

#### HallsofIvy

MHF Helper
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.