Can $\sin (x)$ be considered zero at the point at $\infty$ in some context?

It is understood that $\lim_{x \to\infty} \sin (x)$ is not defined, but I have tricked myself into thinking it might be identifiable as zero at the point at infinity after thinking about continuity, differentiability, and symmetry and then reading J.G.'s answer to why there is a factor of $2\pi$ in $\delta(k)=\frac{1}{2\pi}\int_{-\infty}^\infty \exp\left(ikx\right)dx$.

Let's start with J.G.'s answer to the $\delta$-function question.

\begin{equation}\tag{1}\label{eq1}\int_{-\frac{1}{\epsilon}}^\frac{1}{\epsilon}e^{ikx}dx=\frac{1}{ik}\left(e^{ik/\epsilon}-e^{-ik/\epsilon}\right)=\frac{2}{k}\sin\left(\frac{k}{\epsilon}\right)\end{equation}

Assuming k is nonzero, the integral in \ref{eq1} is supposed to be zero in the $\lim_{\epsilon\to 0^+}$, otherwise the definition of Dirac delta would need to be changed (please correct me if I am wrong).

But would this not then imply that:

\begin{equation}\tag{2}\label{eq2}\lim_{\epsilon \to 0^+,k\neq0} \frac{2}{k}\sin\left(\frac{k}{\epsilon}\right)=0\end{equation}

Now, I have also convinced myself that $\sin(x)$ might be zero at the point at infinity using ideas of continuity, differentiability, and symmetry.

$\sin (x)$ is odd in $x$ so $\sin (x) = \sin(-x)\Rightarrow \sin(x)=0\Rightarrow x=n\pi $ and so $x$ and $-x$ are separated by $2n\pi$ which is a whole number of wavelengths which means we can "wrap the function around and connect the two ends together" and the shape would be continuous and differentiable. The idea of a function being continuous and differentiable at the point at infinity makes so much sense.

So I am wondering if there is any context where $\sin(x)$ can be identified as zero at the point at infinity.

Now, though I did not mention $\cos(x)$ in the question, I would appreciate commentary on that as well.

Since $\cos(x)=\sin\left(x-\frac{\pi}{2}\right)$, one would think that if all the above ideas about $\sin(x)$ were true, then $\cos(x)$ would be $\pm 1$ at the point at $\infty$ ( you could also use continuity, differentiability, and symmetry for $\cos$ on its own ). Either way (plus or minus) the area under the curve is zero and the idea that there should/should not be be a whole number of wavelengths between zero and $\infty$ does not seem relevant enough to decide between plus/minus 1. So if anyone finds a trick which supports $\left|\lim_{x\to\infty}\cos(x)\right|=1$ and/or suggests the sign should be one way or another, that would be much appreciated.