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Let $f : D\subseteq\mathbb{R} \to \mathbb{R}$ be a function. Then, the definition of limit says that: $$\lim_{x \to c} f(x) = L \Longleftrightarrow \forall \varepsilon\in\mathbb{R}:\exists\delta\in\mathbb{R}:\forall x \in D\setminus\{c\} : |x - c| < \delta \longrightarrow |f(x) - L| < \epsilon.$$ Moreover, whenever $\lim_{x \to c} f(x)$ exists, its value $L$ is unique.

Now, let $f : \{c\}\to\mathbb{R}$ be a function whose domain $D$ is a singleton set containing a particular real number $\{c\}$. Then, $\lim_{x\to c}f(x)$ exists and its value is not unique because $\forall x \in \{c\}\setminus\{c\} = \emptyset$ in the following formula is vacuously true for any $L$: $$\forall \varepsilon\in\mathbb{R}:\exists\delta\in\mathbb{R}:\forall x \in \{c\}\setminus\{c\} : |x - c| < \delta \longrightarrow |f(x) - L| < \epsilon.$$

How to fix this situation? By stating in the epsilon-delta definition of limit that an empty domain and a singleton domain are not allowed to have limit because a metric space $D$ must have at least two members?

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  • $\begingroup$ The standard way to avoid this is to only define limits for certain sets $D$, for example ones containing a punctured neighbourhood of $c$. There's no sensible way to define the limit as $x \to c$ of a function $f:\{c\}\to \mathbb{R}$ if you adopt the usual convention that $\lim_{x \to c} f(x)$ should be independent of $f(c)$. $\endgroup$ – Matthew Towers Nov 24 '15 at 9:20
  • $\begingroup$ @mt_: You mean by requiring in the epsilon-delta definition of limit that $D$ must satisfy $(a, c) \bigcup\,(c, b) \subseteq D$ where $a < c < b$? $\endgroup$ – Tadeus Prastowo Nov 24 '15 at 9:35
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    $\begingroup$ Yes, or just requiring $D$ has $c$ as a limit point, or that it contains $(a,c)$ for a one-sided limit -- it depends on the applications and the people the definition is being taught to. $\endgroup$ – Matthew Towers Nov 24 '15 at 9:41
  • $\begingroup$ @mt_: Let $f : [0, \infty) \to \mathbb{R}$ be defined as $f(x) = \sqrt{x}$. Then, $\lim_{x\to 0}\sqrt{x} = 0$ is correct because 0 is a limit point. Now, I have seen a Calculus textbook asserting that $\lim_{x\to c} f(x) = L \Longleftrightarrow \lim_{x\to c^-} f(x) = L = \lim_{x\to c^+} f(x)$. However, for $f(x) = \sqrt{x}$, $\lim_{x\to c^-} f(x)$ fails to exist, and therefore, $\lim_{x\to c} f(x)$ does not exist as well. But, I am sure I am wrong. What is the true definition of $\lim_{x\to c^-} f(x)$ (i.e., one-sided limit)? $\endgroup$ – Tadeus Prastowo Nov 24 '15 at 10:36
  • $\begingroup$ There isn't a "true definition," rather there are many different ones appropriate to different contexts. I agree that a reasonable definition of one-sided limit that applies to functions $f:[0,\infty) \to \mathbb{R}$ should say $\lim_{x \to 0^+}f(x)=0$ in the example you gave. Possibly the calculus text is assuming $f$ is defined on a neighbourhood of $c$ (then that iff result would be true). $\endgroup$ – Matthew Towers Nov 24 '15 at 12:31
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To define the limit of a function $f:E\rightarrow Y$ at a point $p\in E$, $p$ needs to be a limit point of $E$. However, if $E=\{p\}$, we can't define the limit of $f$ at $p$ because $p$ is not a limit point of $\{p\}$.

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  • $\begingroup$ And when the domain $E$ of $f$ is an empty set instead, we say that no limit exists because no point $p$ is in $E$? $\endgroup$ – Tadeus Prastowo Nov 24 '15 at 9:32
  • $\begingroup$ If E is empty, then it's not even possible to think about defining the limit of f at a point p in E. $\endgroup$ – JDZ Nov 25 '15 at 3:49

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