Epsilo-Delta proof - Sin(1/x)

[michonamona]

Homework Statement

Prove that the $$lim_{n\rightarrow0}sin(\frac{1}{x}) = DNE$$

Homework Equations

The Attempt at a Solution

In this case we have:

For every epsilon there exists a delta such that:

$$|x|<\delta$$ implies $$|sin(\frac{1}{x})|<\epsilon$$

I know that $$|sin(\frac{1}{x})|\leq1$$. I'm not sure how to proceed after this point.

I appreciate your help.

M

 

[Count Iblis]

The condition you wrote applies when the limit is zero. If the limit is L, then for every epsilon>0 there exists a delta such that for
|x| < delta we have |sin(x) - L| < epsilon.

Now what do you know about the range of values the function
sin(1/x) will attain for x in the interval 0< |x| < delta for some arbitrary small delta?

 

[michonamona]

The condition you wrote applies when the limit is zero. If the limit is L, then for every epsilon>0 there exists a delta such that for
|x| < delta we have |sin(x) - L| < epsilon.

Now what do you know about the range of values the function
sin(1/x) will attain for x in the interval 0< |x| < delta for some arbitrary small delta?

Ok, I understand that |sin(1/x)| <= 1. So fo some arbitrarily small delta that values of sin(1/x) will be between -1<=sin(1/x)<=1. What can I take away from this?

Thank you

M

 

[Count Iblis]

Ok, I understand that |sin(1/x)| <= 1. So fo some arbitrarily small delta that values of sin(1/x) will be between -1<=sin(1/x)<=1. What can I take away from this?

Thank you

M

And all these values between -1 and 1 will actually be attained. The existence of a limit L, however, requires that you can specify some arbitrary small interval of size epsilon containing L and then for all x smaller than some small delta, the function will only attain values in that small interval. If the limit exist then, in general, the smaller you choose epsilon, the smaller the delta will be that you need for this to work. But the point is that you can always find the delta, no matter how small you choose epsilon.

 

[michonamona]

And all these values between -1 and 1 will actually be attained. The existence of a limit L, however, requires that you can specify some arbitrary small interval of size epsilon containing L and then for all x smaller than some small delta, the function will only attain values in that small interval. If the limit exist then, in general, the smaller you choose epsilon, the smaller the delta will be that you need for this to work. But the point is that you can always find the delta, no matter how small you choose epsilon.

In this case, therefore, it won't work because no matter how small we make our delta around x, sin(1/x) will vacillate from 1 to -1. It will never get to a single limit point. Is this correct?


This is intuitive, but the question is how can I translate this mathematically? So going back to my original question, I'm stuck in$$|sin(\frac{1}{x})|\leq1$$ how do I proceed to complete the proof?

I appreciate all of your help,

M

 

[Count Iblis]

Yes, that's correct. What you do is you assume that the limit exists and is equal to some L. Then you choose epsilon smaller than 1 and show that no value for delta exists such that sin(1/x) for
0<x<delta will always fall within epsilon of L. This thus yields a contradiction.