False notion of Limit in Stewart's "Calculus"

Question

In calculus, when we take limits of functions, say $\lim_{x\to a}f(x),$ do we require that $x$ tends to $a$ from within the domain?

For example, I would say $\lim_{x\to 0} \sqrt{x}=0$ since I am under the assumption that by "$\lim_{x\to 0}$" we mean that $x\to 0$ and $x$ is in the domain. However, this is not consistent with Stewart's "Calculus."

So at this level, should I say that $\lim_{x\to 0}\sqrt{x}$ is undefined?

Answer 1

Stewart's Calculus book isn't the greatest for getting all the details right. But the book is fine for an introductory course in Calculus where the interest primarily is to be able to calculate limits. The book does provide motivation for topics like limits. One can discuss if things ought to be stated more precisely. And sometimes greater precision doesn't mean greater clarity.

Given a function $f: D \to \mathbb{R}$ with domain $D$ and $a$ is an accumulation point of $D$, we write that $$ \lim_{x\to a} f(x) = L $$ if $$ \forall \epsilon > 0 \exists\delta>0: x\in D , 0<\lvert x-a\rvert < \delta \Rightarrow \lvert f(x) - L\rvert < \epsilon. $$

Often we don't write the $x\in D$ part, but it is assumed. With this definition is is perfectly fine to write that $$ \lim_{x\to 0} \sqrt{x} = 0. $$

Answer 2

Summarizing our discussion with Jean-Claude Arbaut: It depends.

The definition you give us from Stewart's calculus is what I would call very sloppy but what passes for "easy to understand" in some beginning calculus courses. There might be extra information in the textbook that you haven't copied so I don't want to judge it too harshly. The point I'm trying to make here though is that there isn't nearly enough information to decide what a limit is supposed to be from that text.

The definition of a limit that I am used to from the US is the following.

Let $D\subseteq\mathbb{R}$, $f:\mathbb{D}\rightarrow \mathbb{R}$ be a function, let $a\in \mathbb{R}$ and let there exist a $\gamma\in\mathbb{R}\;\gamma>0$ such that $(a-\gamma,a+\gamma)\setminus \{a\}\subset Dom(f)$. Then we say that $L$ is the limit of $f$ at $a$, we write $$\lim_{x\to a}f(x)=L$$ if $\forall \epsilon\; (\epsilon>0) \exists \delta\; (\delta>0)$ such that $\forall x \;0<|x-a|<\delta \implies |f(x)-L|<\epsilon$.

To be able to talk about a limit of $f$ at a point $a$, some open interval around $a$ (not containing $a$) must be in the domain of $f$. This is the bit of the definition that ensures that $\lim_{x\to 0}\sqrt{x}$ does not exist.

As I found out from Jean-Claude this is not the definition which is standard in France. Instead of requiring that an open interval around $a$ is part of the domain they take the approach that every neighborhood of $a$ (not including $a$) must intersect the domain of $f$. This ensures that enough points around $a$ are in the domain to make the notion of a limit meaningful. In that case $\lim_{x\to 0}\sqrt{x}=0$ and exists.

Altogether you want to make sure you know what the actual limit definition you are using is and then decide accordingly. Chances are that if you are in the US you are using the definition I present above or something equivalent.

EDIT: Fixed definition to point out the function might not be from all of $\mathbb{R}$.