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!

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 $0.4318404$
