# Proving Limit of sin (n!*R*π) for Rational R

• happyg1
In summary: The problem is proving the limit of sin(n!*R*pi) where R is a rational number. The limit is 0, but finding the value of N for a given epsilon has been difficult. However, by including b in the expansion of n!, the sequence converges at the point where n=b. This simplifies the problem and can be solved using algebra.

#### happyg1

Hi,
Here is my dilemma: I am to prove that sin (n!*R*pi) has a limit, where R is a rational number. I rewrite R as a/b and I can see that whenever n=b, every subsequent term will be zero. I have tried to write this out using the definition of a limit, but I can't seem to break it down. I have been looking at these problems for a long time and I am blocked on this one.
CC

For a sequence $$a_n$$, we say that $$a_n \rightarrow L$$ if for every $$\epsilon > 0$$ we can find an $$N(\epsilon)$$ such that $$| a_n - L | < \epsilon$$ for $$n > N(\epsilon)$$.

You have guessed correctly that the limit is 0. Now I give you an $$\epsilon$$ and ask you to tell me where I should start looking so that the terms of the sequence are always closer than $$\epsilon$$ to the limit. Tell me where to look by giving me N. You've already pointed out that the terms of the sequence equal the limit beyond a certain point ...

Last edited:
hey,
I think I got it. I wasn't including my b in the expansion of the n! as the spot where the sequence converges. I couln't relate the epsilon to the b or the n. It's just the algebra. That's what was giving me the headache. I had been doing the ones that are all polynomials and my brain was fried.

CC

## 1. What is the limit of sin(n!*R*π) for rational R?

The limit of sin(n!*R*π) for rational R is 0. This means that as n approaches infinity, the value of sin(n!*R*π) will approach 0.

## 2. How is the limit of sin(n!*R*π) for rational R proven?

The limit of sin(n!*R*π) for rational R is proven using the squeeze theorem. This theorem states that if a function is "squeezed" between two other functions whose limits both approach the same value, then the original function's limit will also approach that value.

## 3. Can the limit of sin(n!*R*π) for rational R be proven using other methods?

Yes, the limit of sin(n!*R*π) for rational R can also be proven using the definition of a limit, epsilon-delta proofs, and Taylor series expansions.

## 4. Is the limit of sin(n!*R*π) for rational R always 0?

No, the limit of sin(n!*R*π) for rational R is only 0 when R is a rational number. If R is an irrational number, the limit may not exist or may be a different value.

## 5. What is the significance of proving the limit of sin(n!*R*π) for rational R?

Proving the limit of sin(n!*R*π) for rational R is important in mathematics because it helps us understand the behavior of this function as n approaches infinity. It also has applications in fields such as physics, engineering, and statistics.