Given the function
[tex] \dfrac{\sqrt{x^2+1}}{x} [/tex]
The two points concerning the domain are:
Nevertheless, the content of the square root is [tex] x^2+1[/tex], which is always positive, so it's not a problem.
The denominator, which is x, can't be zero, so we can't have [tex] x = 0 [/tex]
So, the domain of the function is
[tex] D = \{ x \in \mathbb{R}:\ x \neq 0\} [/tex]