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!

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
$$
\lim_{x\to\infty\\x \in \mathbb N} x\sin\left(2\pi\sqrt{1+x^2}\right)
 = \pi .
$$
