Finding $\lim _{n\to \infty}n\sin (2\pi \sqrt{1+n^2})$ Question:
Find $\lim \limits _{n\to \infty}n\sin \left (2\pi \sqrt{1+n^2}\right )$, where $n\in \mathbb{N}$.
Context:
This was a question given in my workbook, where it mentioned a hint that $n$ belonging to natural numbers is an important piece of information.
Approach:
I tried expanding the inner part using binomial expansion and then using Taylor series to expand the $\sin (\cdot )$ out, but I seem to keep hitting an indeterminate form. Help in understanding would be much appreciated!
 A: Note that $\sqrt{x^2+1}$ is very close to $x$.  And $\sin$ has period $2\pi$.
Computations.  As integer $x \to \infty$,
\begin{align}
\sqrt{x^2+1} &= x\sqrt{1+\frac{1}{x^2}}
= x\left(1+\frac{1}{2 x^2} + o(x^{-2})\right)
= x + \frac{1}{2 x} + o(x^{-1})
\\
\sin\left(2\pi\sqrt{x^2+1}\right) &=
\sin\left(2\pi x + \frac{\pi}{x} + o(x^{-1})\right) =
\sin\left(\frac{\pi}{x} + o(x^{-1})\right) = \frac{\pi}{x}+o(x^{-1})
\\
x\sin\left(2\pi\sqrt{x^2+1}\right) &= \pi + o(1)
\end{align}
So we conclude
$$
\lim_{x\to\infty\\x \in \mathbb N} x\sin\left(2\pi\sqrt{1+x^2}\right)
 = \pi .
$$
A: Let $\{y\} = y-\lfloor y \rfloor $
That is, $\{y\}$ is the non-integer portion of $y$.
$x\sin (2\pi\sqrt{x^2+1}) = x\sin (2\pi\{\sqrt{x^2+1}\})$
The binomial expansion on $(x^2+1)^\frac 12 = x + \frac 1{2x} - \frac {1}{8x^2} + \cdots$
Since $x$ is an integer, $\frac {1}{2x} - \frac {1}{8x^2}\le \{\sqrt {x^2+1}\} \le \frac {1}{2x}$
$\lim_\limits {x\to \infty} x\sin (2\pi(\frac 1{2x}-\frac {1}{8x^2}) < \lim_\limits {x\to \infty} x\sin(2\pi\sqrt{1+x^2}) < (x)(2\pi)(\frac 1{2x}) = \pi$
It is easy enough to show that the limit on the far left approaches $\pi$ as well, squeezing the target limit.
A: The main idea is that $\sqrt{1+n^2}\approx n$ is big.  Using Taylor expansion of square-root around 1 is not a good idea because convergence deteriorates as $n\to\infty$. Thus:
$$\begin{align}
n\sin\big(2\pi\sqrt{1+n^2}\big)
&= n\sin\big(2\pi n\sqrt{1+1/n^2}\big) \\
&= n\sin\big(2\pi n\sqrt{1+1/n^2}-2\pi n\big) \\
&= n\sin\Big(2\pi n\big(\sqrt{1+1/n^2}-1\big)\Big) \\
&= n\sin\big(\pi /n + o(n^{-3})\big) \\
\end{align}$$
Where in the last step we finally used Taylor of square-root: $\sqrt{1+x^2}=1+x^2/2+o(x^4)$.  Then
$$\begin{align}
\lim_{n\to\infty} n\sin\big(2\pi\sqrt{1+n^2}\big)
&= \lim_{n\to\infty} n\sin\big(\pi /n + o(n^{-3})\big) \\
&= \lim_{x\to0+} \frac{\sin\big(\pi x + o(x^3)\big)}x \\
&= \pi
\end{align}$$
A: As said in comment and answers $$\sin \left(2 \pi  \sqrt{n^2+1}\right)=\sin \left(2 \pi  \sqrt{n^2+1}-2 \pi  n\right)$$
When $n$ is large
$$\sqrt{n^2+1}-n=\frac{1}{2 n}-\frac{1}{8 n^3}+\frac{1}{16
   n^5}+O\left(\frac{1}{n^7}\right)$$
$$\sin \left(2 \pi  \sqrt{n^2+1}\right)=\frac{\pi }{n}-\frac{\pi  \left(3+2 \pi ^2\right)}{12 n^3}+\frac{\pi 
   \left(15+15 \pi ^2+\pi ^4\right)}{120
   n^5}+O\left(\frac{1}{n^7}\right)$$
$$n\sin \left(2 \pi  \sqrt{n^2+1}\right)=\pi -\frac{\pi  \left(3+2 \pi ^2\right)}{12 n^2}+\frac{\pi  \left(15+15 \pi
   ^2+\pi ^4\right)}{120 n^4}+O\left(\frac{1}{n^6}\right)$$ which, for sure, gives the desired limit.
But, it also gives a simple way to approximate the value of the sine.
Just use $n=7$; the truncated series gives, as an approximation,
$$\sin \left(10 \sqrt{2} \pi \right) \sim \frac{\pi  \left(286665-965 \pi ^2+\pi ^4\right)}{2016840}=0.4318486$$ while the exact value is $0.4318404$
