# Domain of definition of $\frac{x}{\sin(x)}$.

Would the domain of definition of $\displaystyle\frac{x}{\sin(x)}$ be $0$? The function on a graphing calculator looks like it has a lot of infinite limits, so I'm not sure.

Let $f$ and $g$ be two real-valued functions with domains $\mathcal{D}(f)$ and $\mathcal{D}(g)$ respectively. The quotient function $\frac{f}{g}$ has domain
$$\mathcal{D}\left(\frac{f}{g}\right) = \{x\in \mathcal{D}(f)\cap\mathcal{D}(g)\ |\ g(x) \neq 0\}.$$
In this case $f(x) = x$, $g(x) = \sin x$, and $\mathcal{D}(f) = \mathcal{D}(g) = \mathbb{R}$, so $\mathcal{D}(f)\cap\mathcal{D}(g) = \mathbb{R}$. As $\sin x = 0$ precisely when $x = k\pi$ for $k \in \mathbb{Z}$, we need to exclude these points from the domain. Therefore, the domain of $\displaystyle\frac{x}{\sin x}$ is $\mathbb{R}\setminus\{k\pi\ |\ k \in \mathbb{Z}\}$.
• I'd delete my answer if you were to just mention that you're restricting $\sin$ to $\Bbb R$, though it's clear that is the case. – Git Gud Feb 24 '13 at 12:08
• The work of Michael Albanese is good, but we can extend this function at $0$ so the domain become $\mathbb{R}\setminus\{k\pi|k\in\mathbb{Z}^*\}$. – user63181 Feb 24 '13 at 12:32
• @sbr: What you are referring to is the fact that $\frac{x}{\sin x}$ has a removable discontinuity at $x = 0$, and hence we can continuously extend the function to this point. That is true, but beside the point. Your proposed answer is not the correct answer to the question at hand. I could extend the function to all of $\mathbb{R}$ if I wanted to (albeit, not continuously), but that does not change the domain of the function $\frac{x}{\sin x}$. – Michael Albanese Feb 24 '13 at 13:20
No, domain of $\frac x{\sin x}$ doesn't include points where $\sin x = 0$, since denominator becomes $0$ in this case. $$\sin x = 0 \\ x = \pi k, \quad k \in \mathbb Z$$ so domain is $D[f] = \mathbb R\setminus \bigcup_{k=-\infty}^{k = +\infty}\{\pi k\}$