Reconciling Rudin's limit definitions & one-sided limit uniqueness When $f : [0, \infty) \to \mathbb{R}$ defined as $f(x) = \sqrt{x}$, based on Rudin's definitions of limit and one-sided limits, I can assert that the following limit exists because 0 is a limit point: $$\lim_{x\to 0} f(x) = 0,$$
the following right-hand limit exists because $f$ is defined on $(0, 1)$ and for all sequences $\{t_n\}$ in $(0, 1)$ such that as $n \to\infty$, $t_n\to 0$, $f(t_n) \to 0$: $$f(0+) = 0,$$
but the following left-hand limit does not exist because $f$ is not defined on $(-1, 0)$ and for all sequences $\{t_n\}$ in $(-1, 0)$ such that as $n \to\infty$, $t_n\to 0$ but $f(t_n)\nrightarrow 0$ due to $f$ being undefined on $(-1, 0)$: $$f(0-)\text{ does not exist}$$
Now based on Rudin assertion, the following holds: $$\lim_{x\to 0}\sqrt{x} = 0 \Longleftrightarrow f(x+) = f(x-) = \lim_{x\to 0} \sqrt{x} = 0\quad\quad\text{(Assertion 1)}$$
However, since the left-hand limit does not exist, the statement does not hold, and therefore, $\lim_{x\to 0}\sqrt{x}$ does not exist. But, this contradicts the earlier definition that $\lim_{x\to 0}\sqrt{x}$ indeed exists because 0 is a limit point.
To resolve that, the comment to my question here says that there is no sequence $\{t_n\}$ in $(-1, 0)$ such that as $n\to\infty$, $t_n\to 0$ and $f(t_n) \to 0$. And, because Rudin requires for all sequences, the non-existence of such a sequence makes the condition vacuously true, and consequently, the left-hand limit exists.
Now, since the left-hand limit exists, it is supposed to be 0 (i.e. $f(0-) = 0)$. But, I can also argue that the left-hand limit is any value other than 0, for example $f(0-) = 7$, because of the vacuous truth. Hence, Assertion 1 fails to hold because the left-hand limit is not unique. This again contradicts the earlier definition that $\lim_{x\to 0}\sqrt{x}$ indeed exists because 0 is a limit point.
How to resolve this contradiction?
 A: Rudin's Definition 4.1:

Let $X$ and $Y$ be metric spaces; suppose $E \subseteq X$, $f$ maps $E$ into $Y$, and $p$ is a limit point $E$. We write $f(x)\to q$ as $x\to p$, or $$\lim_{x\to p} f(x) = q$$ if there is a point $q \in Y$ with the following property: For every $\varepsilon > 0$ there exists a $\delta > 0$ such that $$d_Y(f(x), q) < \varepsilon$$ for all points $x\in E$ for which $$0 < d_X(x, p) < \delta.$$

Rudin's Definition 4.25:

Let $f$ be defined on $(a, b)$. Consider any point $x$ such that $a \leq x < b$. We write $$f(x+) = q$$ if $f(t_n) \to q$ as $n\to\infty$, for all sequences $\{t_n\}$ in $(x, b)$ such that $t_n\to x$. To obtain the definition of $f(x-)$, for $a < x \leq b$, we restrict ourselves to sequences $\{t_n\}$ in $(a, x)$. It is clear that any point $x$ of $(a, b)$, $\displaystyle\lim_{t\to x} f(t)$ exists if and only if $$f(x+) = f(x-) =\lim_{t\to x} f(t).$$

Hence, it is clear that Definition 4.1 is more general than Definition 4.25 that requires $f$ to be defined on an interval $(a, b)$. Moreover, based on Definition 4.25, the assertion that $\displaystyle\lim_{t\to x} f(t)$ exists if and only if $$f(x+) = f(x-) = \lim_{t\to x} f(t)$$ only holds when $x \in (a, b)$. When $x = a$ or $x = b$, Definition 4.25 asserts nothing, but Definition 4.1 can assert that $\lim_{x\to p}f(x) = q$.
To conclude, Definition 4.1 and Definition 4.25 can be reconciled in the following way guaranteeing the uniqueness of one-sided limit:

Let $X$ and $Y$ be metric spaces; suppose $E \subseteq X$, $f$ maps $E$ into $Y$, and $p$ is a limit point $E$. We write $f(x)\to q$ as $x\to p$, or $$\lim_{x\to p} f(x) = q$$ if there is a point $q \in Y$ with the following property: For every $\varepsilon > 0$ there exists a $\delta > 0$ such that $$d_Y(f(x), q) < \varepsilon$$ for all points $x\in E$ for which $$0 < d_X(x, p) < \delta.$$ Moreover, if $p \in (a, b)$, then $\displaystyle\lim_{x\to p} f(x)$ exists if and only if $$f(p+) = f(p-) =\lim_{x\to p} f(x).$$ where we write $$f(x+) = q$$ for any point $x$ such that $a \leq x < b$ if $f(t_n) \to q$ as $n\to\infty$, for all sequences $\{t_n\}$ in $(x, b)$ such that $t_n\to x$, while when we write $f(x-)$, for $a < x \leq b$, we restrict ourselves to sequences $\{t_n\}$ in $(a, x)$.

In light of the reconciled definitions, the assertion in the comments of this post that it is 100% not sloppy to write $\lim_{x\to 0}$ when the domain is $\{x\,|\,x\geq 0\}$ is correct because $0$ is not an interior point. As for the question posted here and here, the logical fallacy is to not recognize that it is invalid to discuss the assertion $\displaystyle\lim_{t\to x} f(t)$ exists if and only if $$f(x+) = f(x-) = \lim_{t\to x} f(t)$$ when $x$ is not an interior point.
