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

Question

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 $a$.

Answer 1

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 $\lim_{x\to a} f(x) = L : \iff$

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

Answer 2

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.

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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.

Answer 3

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 $0$.