Famously, $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x} = 1.$ However, whether or not it is considered continuous at $x = 0$ seems unclear to me. Another post has pointed out that Rubin's Principles of Mathematical Analysis writes
If $x$ is a point in the domain of definition of the function $f$ at which $f$ is not continuous, we say that $f$ is discontinuous at $x$, or that $f$ has a discontinuity at $x$.
It does not mention discontinuity at points not in the domain of $f$, implying $f:D \to \mathbb{R}$ is only locally discontinuous at point $a\leftarrow a\in D.$ Wolfram Mathworld supports that, saying
...The given definition of removable discontinuity fails to apply to functions $f$ for which $\displaystyle\lim_{x\to x_0}f(x)=L$ and for which $f(x_0)$ fails to exist; in particular, the above definition allows one only to talk about a function being discontinuous at points for which it is defined.
Then, 1) Would $\dfrac{\sin x}{x}$ be considered neither continuous nor discontinuous at $x = 0$ since its domain is $\mathbb{R}~\backslash~\{0\}$?
Other definitions such as here say that
A function $f (x)$ is continuous at $x = a$ if the following all hold: $f(a)$ exists and $\displaystyle\lim_{x\to a} f(x)$ exists s.t. $\displaystyle\lim_{x\to a} f(x) = f(a)$,
which classifies $x = 0$ for $\dfrac{\sin x}{x}$ as a removable discontinuity. If we look to the rigorous $\epsilon-\delta$ definition of continuity, it states that for $f$ to be continuous at $x = a$, $\forall \epsilon > 0 \exists \delta>0$ s.t. $|x-a|<\delta,~x\neq a \implies |f(x)-f(a)|<\epsilon.$ Since the definition includes $f(a)$, which is undefined, is that considered discontinuous?
Which definition is correct?
