I wrote the definitions below. Are they accurate? If not, what correction(s) should be made? You may think based on an apparent lower bound for my level of mathematical maturity that I could answer this question for myself, but peer review is important to me.

The $\varepsilon$-$\delta$ definition of a limit. Let $f$ be a function defined on an open interval containing a number $a$ except possibly at $a$. Then, for some number $L$, \begin{equation*} \lim_{x \rightarrow a} f\left(x\right) = L \end{equation*} if $\forall \varepsilon > 0, \exists \delta > 0$ such that $\forall x$ in the domain of $f$, \begin{equation*} |x - a| < \delta \implies |f\left(x\right) - L| < \varepsilon. \end{equation*} The $\varepsilon$-$\delta$ definition of a left-hand limit. Let $f$ be a function defined on an interval with a number $a$ as its right endpoint. Then, for some number $L$, \begin{equation*} \lim_{x \rightarrow a^-} f\left(x\right) = L \end{equation*} if $\forall \varepsilon > 0, \exists \delta > 0$ such that $\forall x$ in the domain of $f$, \begin{equation*} a - \delta < x < a \implies |f\left(x\right) - L| < \varepsilon. \end{equation*}

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  • $\begingroup$ seems correct to me $\endgroup$ – Deepesh Meena Aug 4 '18 at 19:19
  • $\begingroup$ Because I've never seen limits defined similarly enough to how I defined them to feel safe with how I defined them. Why else might one pose such a question? Though it may not be appropriate for me to say so, I recognize that I shouldn't have taken offense to what you said and retaliated; sorry. $\endgroup$ – PiKindOfGuy Aug 4 '18 at 21:00
  • $\begingroup$ No worries and thanks for the apology. I've removed my earlier comment. What references do you have for these different definitions. $\endgroup$ – Rob Arthan Aug 4 '18 at 21:47
  • $\begingroup$ Google results when searching for the definition of a limit and the definition of a left-hand limit. I also have (somewhat) stored in memory what I've read in books, what I've seen on Khan Academy and what I wrote down during my first calculus course. Why? $\endgroup$ – PiKindOfGuy Aug 4 '18 at 22:03
  • $\begingroup$ I wrote "my own" definitions because I didn't like (enough) how other definitions were written and I feared that in doing so I might've messed something up. $\endgroup$ – PiKindOfGuy Aug 4 '18 at 22:06

For the first one, I think it's more common to use the definition with $0 < |x-a| < \delta$ instead of $|x-a|<\delta$, although you can find both variants in different sources. In other words, even if $f(a)$ happens to be defined, that value shouldn't be taken into account as far as the limit is concerned. (With your definition, the limit must equal $f(a)$ if $f(a)$ is defined.)

Also, requiring $f$ to be defined on a punctured interval around $a$ is a bit strong; it's enough that $a$ is a limit point of the domain of $f$.

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  • $\begingroup$ In response to your first paragraph: I prefer simplicity (with accuracy) over clarity. $\endgroup$ – PiKindOfGuy Aug 4 '18 at 20:30
  • $\begingroup$ In response to your second paragraph: though f needs to be defined on a punctured interval (in my definition), that doesn't mean that within f's domain the interval is punctured. See my related response to José Carlos Santos's answer. Is there still an issue here with my definition? $\endgroup$ – PiKindOfGuy Aug 4 '18 at 20:35
  • $\begingroup$ Hans' comment bears thinking about, and I think that there's a good case to define "continuous at $a$" first, and then say that $\lim_{x \to a} f(x) = L$ exactly if $g(x) = \begin{cases}f(x) & x \ne a \\ L & x = a \end{cases}$ is continuous at $a$. (This requires that you fuss a little about whether the definition of "continuous at $a$" should require that the domain of the function contain some interval around $a$, however.) $\endgroup$ – John Hughes Aug 4 '18 at 20:36
  • $\begingroup$ @PiKindOfGuy: I don't understand what you want to say with those comments. $\endgroup$ – Hans Lundmark Aug 5 '18 at 8:22

It is almost totally correct. The $\varepsilon-\delta$ part is fine. The only problem is that the definition of left-hand limit doesn't requere that $a$ is the right endpoint of the domain of $f$.

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    $\begingroup$ I'm not saying that a is the right endpoint of the domain, only that the function is defined on some interval with a as a right endpoint. The function's domain can be larger than the interval. Am I misunderstanding the meaning of the words I'm using? $\endgroup$ – PiKindOfGuy Aug 4 '18 at 19:23
  • $\begingroup$ I agree. The left hand limit as $x \to 1$ of $f(x) = 2x$ (defined on $\Bbb R$) as is $2$, for instance, even though $x = 1$ is not the right-hand end of the domain of $f$. $\endgroup$ – John Hughes Aug 4 '18 at 19:26
  • $\begingroup$ @JohnHughes See my comment $\endgroup$ – PiKindOfGuy Aug 4 '18 at 19:31
  • $\begingroup$ I meant to say that I agreed with your comment, @PiKindOfGuy, rather than with Jose (with whom I usually agree!). $\endgroup$ – John Hughes Aug 4 '18 at 19:36
  • $\begingroup$ @JohnHughes Sorry for my confusion and thanks for the input. Jose's answer is still valuable because it highlights that "defined on an interval..." may---to some---mean the same as "defined only on an interval..." $\endgroup$ – PiKindOfGuy Aug 4 '18 at 19:41

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