Modifying definition of limit so function isn't defined around $a$

This is a kinda follow-up question to this question here.

My book (Stewart's Calculus) says that the definition of $$\lim_{x\to a} f(x) = L$$ is that

Let $f$ be a function defined on some open interval that contains the number $a$, except possibly at $a$ itself. Then we say that the limit of $f(x)$ as $x$ approaches $a$ is $L$, and we write $\lim_{x\to a} f(x) = L$ if for every number $\epsilon > 0$ there is a number $\delta > 0$ such that if $0 < \lvert x - a \rvert < \delta$ then $\lvert f(x) - L\rvert < \epsilon$.

With this definition it wouldn't be true that $\lim_{x\to 0} \sqrt{x} = 0$. But from the earlier question some would say that it is fine to say that $\lim_{x\to 0} \sqrt{x} = 0$ and one doesn't have to use a right hand limit.

My question is how to modify the definition given in my book so that it allows one to say that $\lim_{x\to 0 }\sqrt{x} = 0$.

My suggestion is simply dropping the requirement that the function be defined around $a$.

• Usually, for a given function $f$ of natural domain $\mathcal{D}_f$ the definition of the limit is : $$\forall \epsilon > 0, \exists \delta > 0/ \color{red}{\forall x \in \mathcal{D}_f}, |x-a| < \delta \Rightarrow |f(x)-L|<\epsilon$$ In the case of the square root function, as $\mathcal{D}_f$ is $\mathbb{R}_+$, it makes the definition valid again. – Wyllich Aug 30 '17 at 14:04
• You shoul write $0<|x-a| < \delta$ instead of $|x-a| < \delta$ – Fred Aug 30 '17 at 14:11
• @Fred Does that mean $\lim\limits_{x \to 0} \delta_0 = 0$ ? I thought it would be undefined... Good too know. – Wyllich Aug 30 '17 at 14:19
• What is $\delta_0$ ?? – Fred Aug 30 '17 at 14:20

Defintion: let $D \subset \mathbb R$ non void , $f:D \to \mathbb R$ a function and let $a$ be an accumulation point of $D$. Then $\lim_{x\to a} f(x) = L : \iff$

$\forall \epsilon > 0, \exists \delta > 0: \forall x \in D, 0<|x-a| < \delta \Rightarrow |f(x)-L|<\epsilon$.

• What is a limit point? – John Doe Aug 30 '17 at 14:47
• OOPs a misprint. I mean "accumulation point" , sorry ! – Fred Aug 30 '17 at 14:50
• BTW a "limit point" is same as an "accumulation point" and it takes fewer letters to type. Various authors may prefer one term over another. – Paramanand Singh Sep 5 '17 at 17:00

Just replace the definition of your textbook by the following:

Definition 2: Let $\emptyset\neq A\subseteq \mathbb {R}$ and $f:A\to\mathbb{R}$ be a function. Further let $a$ be an accumulation point (or a limit point) of $A$. A real number $L$ is said to be the limit of $f$ at $a$ (denoted by $\lim\limits_{x\to a} f(x) =L$ or $f(x) \to L$ as $x\to a$) if for any given real number $\epsilon >0$ there is a corresponding real number $\delta>0$ such that $|f(x) - L|<\epsilon$ whenever $0<|x-a|<\delta$ and $x\in A$.

For expression $\lim_{x\to a} f(x)$ to be defined it is an important pre-requisite that $a$ be an accumulation point of the domain of $f$. The cases when $a=\pm\infty$ are handled differently.

I prefer the definition given in your question compared to the one provided above because it avoids the term "accumulation point" and is suitable for a first course in calculus. Moreover the second definition does not offer anything extra as far any theorems in introductory calculus course are concerned.

The devil is in the details of how you define your function. The fact that $f$ is defined on an open interval is important here.

A function is two things, a domain and a "rule." You have specified the rule, that $x$ maps to $\sqrt x$, but not the domain. Typically, when the domain is unclear, we pick the largest domain that makes sense. In this case, $[0,\infty)$. Since we can't take the square root of negative numbers in real analysis, it is impossible to have $\sqrt x$ defined on an open interval containing $0$.

This point is a bit tricky and I see your confusion. Here is a clearer way to write the first line of your definition: "let $I$ be an open interval containing $a$, and let $f$ be a function defined on $I\setminus \{a\}$." In this version, it is clear that $(0,1)$ doesn't work since it doesn't contain $0$ at all. Any attempt to "capture" zero in an open interval will necessarily also contain some negative numbers, and we cant do that here.

There are two ways around this: some books will introduce one-sided limits as their own concept. I find it more intuitive to do the following: take your open interval $I$ and the function $f$ defined on $I\setminus\{a\}$. Then only talk about the values in $I\setminus\{a\}$ which make sense to talk about. In other words, we look at the values of $x$ in the intersection of $I\setminus\{a\}$ and the domain of $f$.

This way, we can look at the interval, for example, $(-1,1)$. Then it is certainly true that, for $\epsilon>0$, we can find $\delta>0$ for which $|x|<\delta$ implies $\sqrt x<\epsilon$.

Dissect that line one more time: for any values of $x$ satisfying $|x|<\delta$, and for which $f$ is defined at $x$, we have $\sqrt x<\epsilon$. Thus $f$ is indeed continuous at $0$.