# Proving Limit of sin(1/x) Does Not Exist

• sutupidmath
In summary, In this conversation, the following was proved: lim sin(1/x) when x->0 does not exist. i used Haines definition of the limit to prove this. I found two sequences x=1/n*pi, so when n-> infinity x->0, and the other sequence that also converges to zero that i used in this case is x=2/(4n+1)*pi, this also when n-> infinity then x->0. now when we take take the corresponding sequences we get: lim sin n*pi, n-> infinity, and lim sin (4n+1)*pi/2, as n-> infinity, the problem is here, i
sutupidmath
I have to prove that lim sin(1/x) when x->0 does not exist. i used Haines definition of the limit to prove this. I found two sequences x=1/n*pi, so when n-> infinity x->0, and the other sequence that also converges to zero that i used in this case is x=2/(4n+1)*pi, this also when n->infinity then x->0. now when we take take the corresponding sequences we get:

lim sin n*pi, n-> infinity, and lim sin (4n+1)*pi/2, as n-> infinity,

the problem is here, i am not sure at this case if i can take lim sin n*pi, n-> infinity =0 and lim sin (4n+1)*pi/2, as n-> infinity= 1, i think i should do like this. But at this case what i am courious to know is why should i take these results??

any help would do.

Assume it does converge to a limit as x->0.

Now, let the epsilon be 0.1, then we should be able to find a gamma that satisfies this condition by the definition of the limit.

Can you see a problem that occurs when e=0.1?

What exactly is "Haines definition of the limit"?

HallsofIvy said:
What exactly is "Haines definition of the limit"?

His professors name?

You have a sequence x_n = sin npi, so that x_n=0 for all n.
You have a sequence y_n = sin (4n+1)pi/2, so that y_n=1 for all n.

They obviosuly converge, since they are constant. That x_n=sin npi is neither here nor there.

yeah matt grime i do understand that x_n = sin npi, so that x_n=0 for all n, and also y_n = sin (4n+1)pi/2, so that y_n=1 for all n. But my question is :

If it is still safe to jump to these conclusions when n->infinity?

thnx

Of course it is. Why wouldn't it be? x_n is always zero. Irrespective of what n is. Do you think that at some point sin(npi) is going to stop being zero?

Actualy scrub that. Let me put it this way. What has n going to infinity got to do with the value of sin(npi)? You appear to be thinking of things in the wrong order.

taking the limit of sin(npi) as n tends to infinity does not affect the values sin(npi) at all. How can it. They're fixed (and always 0).

Last edited:
ej matt grime thank you, this is what i was not quite sure about. Now it is all clear thnx for your help guys

## What is the limit of sin(1/x) as x approaches 0?

The limit of sin(1/x) as x approaches 0 does not exist. This means that no single value can be assigned to the limit as x gets closer and closer to 0.

## How can you prove that the limit of sin(1/x) does not exist?

The limit of sin(1/x) can be proven to not exist by showing that the function oscillates or approaches different values as x approaches 0 from different directions. This can be done through the use of the squeeze theorem or by examining the graph of the function.

## Why is it important to prove that the limit of sin(1/x) does not exist?

Proving that the limit of sin(1/x) does not exist is important in understanding the behavior of this function and its limits. It also has implications in calculus and other areas of mathematics where limits are used.

## What is the difference between the limit of sin(1/x) and the limit of 1/x as x approaches 0?

The limit of sin(1/x) does not exist, while the limit of 1/x as x approaches 0 is equal to 0. This is because the function sin(1/x) oscillates and does not approach a single value as x gets closer to 0, while the function 1/x approaches 0 as x approaches 0.

## Can the limit of sin(1/x) be proven to exist?

No, the limit of sin(1/x) cannot be proven to exist. This is because the function oscillates and does not approach a single value as x gets closer to 0, which is a requirement for the existence of a limit.

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