Determine the following limit: $$\lim_{n\to\infty}n\sin(2\pi\sqrt{1+n^2}).$$
My Approach: I have myself solved the problem over $n\in\mathbb N$, but do not have any clue on how to solve it over $n\in\mathbb R$. My solution for $n\in\mathbb N$ is as follows but my question is is the following problem even solvable over $n\in\mathbb R$? Anyways, here is my solution for $n\in\mathbb N$.
\begin{align*}
\lim_{n\to\infty}n\sin(2\pi\sqrt{1+n^2})&=\lim_{n\to\infty}n\sin(2\pi\sqrt{1+n^2}-2n\pi)\\
&=\lim_{n\to\infty}n\sin(2\pi(\sqrt{1+n^2}-n))\\
&=\lim_{n\to\infty}n\sin\left(2\pi\left(\dfrac{1}{\sqrt{1+n^2}+n}\right)\right)\\
&=\left(\lim_{n\to\infty}\dfrac{\sin\left(\dfrac{2\pi}{\sqrt{1+n^2}+n}\right)}{\dfrac{2\pi}{\sqrt{1+n^2}+n}}\right)\cdot\lim_{n\to\infty}\left(n\cdot\dfrac{2\pi}{\sqrt{1+n^2}+n} \right)\\
&=\pi
,\end{align*}
which completes the proof when $n\in\mathbb N$. But the problem is when I used that $-2n\pi$, I used the fact that $n\in\mathbb N$, and nowhere it is told what $n$ is which is why I doubt if this problem is solvable.
Thank you.
Best Answer
Consider the sequences $$ x_n = \sqrt{n^2 - 1} \qquad \text{and} \qquad y_n = \sqrt{\left( n + \tfrac{1}{4} \right)^2 - 1}. $$ For $f(x) = x \sin(2 \pi \sqrt{1 + x^2})$, one has $$ f(x_n) = x_n \sin(2 \pi n) = 0 \qquad \text{and} \qquad f(y_n) = y_n \sin \left( 2 \pi n + \tfrac{\pi}{2} \right) = y_n \to \infty \text{ as } n \to \infty. $$ Thus $\lim_{x \to \infty} f(x)$ does not exist.