# Thread: Limit of sin(2^n) as n-> inifinity

1. ## Limit of sin(2^n) as n-> inifinity

I want to calculate this limit for quite some days now and I have made almost no progress whatsoever.
(limit of sin(2^n) as n-> infinity, neN ) I suspect either lim(sin(2^n)) = 0 or it is non-existing.
Thus fur I have used numerous trigonometry properties resulting to nowhere,
also I tried to prove it converges to 0 but turned out my calculations lead it diverging which is impossible.
I can't seem to find a proper way of calculating/solving this other than an informal theoretical way.

2. Originally Posted by Kiyo
I want to calculate this limit for quite some days now and I have made almost no progress whatsoever.
(limit of sin(2^n) as n-> infinity, neN ) I suspect either lim(sin(2^n)) = 0 or it is non-existing.
Thus fur I have used numerous trigonometry properties resulting to nowhere,
also I tried to prove it converges to 0 but turned out my calculations lead it diverging which is impossible.
I can't seem to find a proper way of calculating/solving this other than an informal theoretical way.
Why that is not acceptable?

How do you calculate limit of an oscillating function?

3. I don't seem to know how to calculate a limit of an oscillating function.
When I said an informal theoretical way I meant a poor explanation, where as in physics an oscillating object, take a string for example, if it oscillates with an incredible amount of angular velocity, the width it has is negligible, thus close to 0.
What I tried to do was prove somehow that |sin(2^n)| is a decreasing function.

4. Originally Posted by Kiyo
I don't seem to know how to calculate a limit of an oscillating function.
When I said an informal theoretical way I meant a poor explanation, where as in physics an oscillating object, take a string for example, if it oscillates with an incredible amount of angular velocity, the width it has is negligible, thus close to 0.
What I tried to do was prove somehow that |sin(2^n)| is a decreasing function. ..... No - value of sin(2^n) will depend on the remainder of (2^n)/(2π).
|sin(2^n)| ≤ 1

5. How would you calculate:

$\displaystyle{\lim_{x \to \infty} \sin(x)}$

6. I would say since -1≤ sin(x) ≤ 1 and x∈R
sin(x) oscillates between all the values within -1,1 thus the

lim(sin(x)) as x->infinity

Does not exist.
But I thought otherwise about sin(2^n) as n
∈N,
it doesn't take all the values between -1,1 and 2^n isn't a linear function.