Question Regarding Existence of One Sided Limits In a Calculus book, I had read the following proposition:

For a function $f:X\to \mathbb{R}$, $X\subseteq \mathbb{R}$ then $\lim\limits_{x\to x_0}f(x)$ exists if and only if $\lim\limits_{x\to x_0^+}f(x)$ and $\lim\limits_{x\to x_0^-}f(x)$ exist and $\lim\limits_{x\to x_0^+}f(x)=\lim\limits_{x\to x_0^-}f(x)$. 

It has come to my attention however that this is wrong.
Let $f:\left[a,b\right] \to \mathbb{R}$, $f(x)=x$. Note that $a$ is an accumulation point (from the right) of the domain of $f$.
Obviously, $\lim\limits_{x\to a^-}f(x)$ does not exist since $a$ is not an accumulation point from the left of the domain of $f$. Using the definition of a limit of a real function (or using the fact that $f$ is continuous on $a$) we can derive that $\lim\limits_{x\to a}f(x)=a$ which is a contradiction to the proposition above.
My question is, in the proposition do we need to additionaly suppose that $x_0$ is an accumulation point from the right and the left of $X$?
If $x_0$ is an accumulation point of $X$ only from the right and $\lim\limits_{x\to x_0^+}f(x)$ exists then is it true that $\lim\limits_{x\to x_0}f(x)$ exists and $\lim\limits_{x\to x_0}f(x)=\lim\limits_{x\to x_0^+}f(x)$?
 A: It's not "wrong"; it depends on your definition of limit.
The calculus textbook is working on a definition of limit that requires the function to be defined on a punctured neighborhood of the point $x_0$. Explicitly, the definition is:

$\lim\limits_{x\to x_0}f(x) = a$ if and only if for every $\epsilon\gt 0$ there exists $\delta\gt 0$ such that for all $x$, if $0\lt |x-x_0|\lt\delta$, then $|f(x)-a|\lt\epsilon$.

The definition implicitly requires there to be a $\delta_0\gt 0$ such that $(x_0-\delta_0,x_0+\delta_0)-\{x_0\}\subseteq\mathrm{dom}(f)$.
Under this definition, the highlighted statement is correct. You cannot compute $\lim\limits_{x\to a}f(x)$ because $f(x)$ is not defined on a punctured neighborhood of $a$.
For a slightly more general definition that requires only $x_0$ to be an accumulation point of the domain, the definition may be modified slightly as follows:

Let $x_0$ be an accumulation point of $\mathrm{dom}(f)$. Then $\lim\limits_{x\to x_0}f(x) = a$ if and only if for every $\epsilon\gt 0$ there exists $\delta\gt 0$ such that for all $x\in\mathrm{dom}(f)$, if $0\lt |x-x_0|\lt\delta$, then $|f(x)-a|\lt\epsilon$.

Under this definition, you can compute limits on endpoints without specifying sides, and the highlighted statement needs to be modified to apply only to points that are accumulation points from both sides.
It's a matter of definition.
