<s>[INST] Evaluating $\lim_{n\to \infty} \sin(\sqrt{n^2+1}\pi)$. (WolframAlpha says it doesn't exist; I get $0$.) I have tried to solve limit, which wolfram says that DNE, but according to my calculations it is equal to 0. Limit is given below
$$\begin{align}
\lim_{n\to \infty} \sin(\sqrt{n^2+1}\pi)
&=\sin(\sqrt{n^2+1}\pi-n\pi+n\pi) \\
&=(-1)^n\sin(\sqrt{n^2+1}\pi-n\pi) \\
&=(-1)^n\sin\left(\frac{(\sqrt{n^2+1}\pi-n\pi)(\sqrt{n^2+1}\pi+n\pi)}{(\sqrt{n^2+1}\pi+n\pi)}\right) \\
&=(-1)^n\sin\left(\frac{n^2\pi^2+\pi^2-n^2\pi^2}{\sqrt{n^2+1}\pi+n\pi}\right) \\
&=(-1)^n\sin\left(\frac{\pi^2}{n\pi(\sqrt{1+\frac{1}{n^2}}+1}\right) \\
&=0 
\end{align}$$
It is because denominator of sin goes to infinity so everything inside sin goes to 0. as we know, sin of that would go to 0 too. And we know that $(-1)^n$ is bounded so we got that bounded * 0 has to be equal to 0.
Am I doing some mistake here ?
 [/INST] The step $\sin(\sqrt{n^2+1}\pi) = (-1)^n\sin(\sqrt{n^2+1}\pi-n\pi)$ holds if $n$ is an integer but does not hold if $n$ is not an integer.
Let's look at this another way.
$\sqrt{n^2+1}$ is unbounded and increasing.  When $n$ is large $\sqrt{n^2+1}\approx n.$
Suppose $n\in \mathbb R$
For large values of $n$ there will be values of $\sin(\sqrt{n^2+1}\pi)$ that equal $1, -1$ and $0$ and the limit does not exist.
e.g. $\sin (\sqrt{100.5^2 + 1}\pi) = 0.9999$
If $n\in \mathbb N$ then for large values of $n$
$\sin(\sqrt{n^2+1}\pi)\approx \sin n\pi = 0$
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