2. ## Re: limit sinus-sqrt

Originally Posted by dhiab
I highly doubt a limit exists. Why would you think it does? I would imagine you can come up with subsequences that could have any limit you desire between 0 and 1.

3. ## Re: limit sinus-sqrt

I agree with SlipEternal. On the other hand we can provide an argument that it's 0. (Clearly there will be other methods that say otherwise.)

Anyway, for very large n, and that $\displaystyle n^2 >> n$, the limit becomes $\displaystyle \lim_{n \to \infty} sin ^2 ( \pi \sqrt{n^2 + n} ) \to \lim_{n \to \infty} sin^2 (\pi n)$. Since n is an integer and $\displaystyle sin(\pi n) = 0$ for all integer values of n then:

$\displaystyle \lim_{n \to \infty} sin ^2 ( \pi \sqrt{n^2 + n} ) \to \lim_{n \to \infty} sin^2 (\pi n) = 0$.

-Dan

4. ## Re: limit sinus-sqrt

We can write
\begin{align} \lim_{n \to \infty} \sqrt{n^2+n} &= \lim_{n \to \infty} \sqrt{\tfrac14(4n^2 + 4n + 1-1)} \\ &= \lim_{n \to \infty}\tfrac12\sqrt{(2n+1)^2 - 1} \end{align}
So clearly, $$\sqrt{n^2+n} \approx \tfrac12 (2n+1)$$ and this is a very slight over-estimate for large $n$. (A better over-estimate is $\sqrt{n^2+n} \approx \tfrac12\left((2n+1)-\frac{1}{2(2n+1)}\right)$).

Thus $$\sin{\big(\pi\sqrt{n^2+n}\big)} \approx \sin {\big(\pi(n+\tfrac12)\big)}$$ and this means that $\sin{\big(\pi\sqrt{n^2+n}\big)}$ alternates between approximately $1$ and approximately $-1$. So there is no limit.

You'll find that my approximation of $\sqrt{n^2+n}$ is closer than topsquark's.

5. ## Re: limit sinus-sqrt

Originally Posted by dhiab
I find it unbelievable that this is even a multiple thread.
LOOK HERE.
It should be clear that there is no limit. Rather the function oscillates between 0 & 1.

6. ## Re: limit sinus-sqrt

I've just realised that I forgot that the $\sin$ function is squared, so the limit is actually $1$.

Plato's comment would make sense for a continuous (real) variable $x$, but not for the discrete (natural) $n$.

7. ## Re: limit sinus-sqrt

More formally, we can demonstrate that $$\frac12\left( (2n+1) - \frac1{2n+1} \right) < \sqrt{n^2 + n} < \frac12(2n+1)$$ and then, perhaps via the identity $\sin{(A-B)} = \sin{(A)}\cos{(B)} - \cos{(A)}\sin{(B)}$, that $$\lim_{n \to \infty} \left|\sin{\left(\pi\sqrt{n^2+n}\right)}\right| = \left| \sin {\left(\tfrac12(2n+1)\pi\right)} \right| = 1$$

8. ## Re: limit sinus-sqrt

Originally Posted by Archie
We can write
\begin{align} \lim_{n \to \infty} \sqrt{n^2+n} &= \lim_{n \to \infty} \sqrt{\tfrac14(4n^2 + 4n + 1-1)} \\ &= \lim_{n \to \infty}\tfrac12\sqrt{(2n+1)^2 - 1} \end{align}
So clearly, $$\sqrt{n^2+n} \approx \tfrac12 (2n+1)$$ and this is a very slight over-estimate for large $n$. (A better over-estimate is $\sqrt{n^2+n} \approx \tfrac12\left((2n+1)-\frac{1}{2(2n+1)}\right)$).

Thus $$\sin{\big(\pi\sqrt{n^2+n}\big)} \approx \sin {\big(\pi(n+\tfrac12)\big)}$$ and this means that $\sin{\big(\pi\sqrt{n^2+n}\big)}$ alternates between approximately $1$ and approximately $-1$. So there is no limit.

You'll find that my approximation of $\sqrt{n^2+n}$ is closer than topsquark's.
Meh. What's a few millions of terms between friends?

-Dan

9. ## Re: limit sinus-sqrt

$1$ in the limit apparently. But I can be picky like that. But you are within an engineer's approximation.

10. ## Re: limit sinus-sqrt

Hello here is a solution