# False notion of Limit in Stewart's “Calculus”

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?

• – DRF Oct 8 '15 at 8:33
• Removed comments: not wrong, but too regionalized to be of any use here, where there are indeed significant differences between french and english usual definitions of a limit. – StayHomeSaveLives Oct 8 '15 at 9:10
• Also deleted comments since it was mostly a discussion. The essential result is you need to figure out what the conventions are where you are taking the class, and if talking with fellow mathematicians make sure your definitions either match up or you know how they are different. – DRF Oct 8 '15 at 9:12
• Even in the U.S., several books define $\lim_{x\to a}f(x),$ just requiring $x$ tends to $a$ from within the domain and NOT requiring that for some $\gamma$, $(a-\gamma,a+\gamma)\setminus \{a\}\subset Dom(f)$. See, for instance, Analysis I, Serge Lang. Another example is: encyclopediaofmath.org/index.php/Limit . According to those US books and sources, it is correct to write $\lim_{x\to 0}\sqrt{x}=0$. In fact, it is completly consistent with writing, for instance, $\lim_{x\to -\infty} e^x=0$. – Ramiro Oct 8 '15 at 13:39

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 $$\lim_{x\to 0} \sqrt{x} = 0.$$

• Did you mean to have $a\in D$? That very much restricts the amount of places you can take limits. And as I asked before what happens to the situation where a function is defined on say just the rationals? I guess the real point is this is not a "mathematics" course but rather a calculus course. That used to annoy me greatly when I started to teach in the US, and apparently it's come back now I'm not teaching there anymore.:D – DRF Oct 8 '15 at 13:04
• This definition strikes me as problematic because it allows for limits at isolated points of a domain. E.g., suppose $D=\mathbb{Z}$ and $a=0$. With $\delta={1\over2}$ regardless of $\epsilon$, the if-then implication $x\in D,0\lt|x-0|\lt{1\over2}\implies|f(x)-L|\lt\epsilon$ is trivially true (for all $x$) because the "if" part is false. The problem with this, I'd say, is that the limit $L$ is not unique. – Barry Cipra Oct 8 '15 at 13:31
• @BarryCipra The definition requires $a$ to be an accumulation point of $D$ (instead of $a\inD$), but the rest of the definition is as Thomas wrote it. (See, for instance: encyclopediaofmath.org/index.php/Limit ). – Ramiro Oct 8 '15 at 13:53
• @Thomas May I suggest that you edit your answer to replace "$a\in D$" by "$a$ is an accumulation point of $D$". Then your definition will match the definition found, for instance, in Analysis I, Serge Lang and in encyclopediaofmath.org/index.php/Limit . Your conclusion, of course, remains the same: it is perfectly fine to write that $$\lim_{x\to 0}\sqrt{x}=0$$. – Ramiro Oct 8 '15 at 14:01
• @Ramiro, I'm fine with that. – Barry Cipra Oct 8 '15 at 14:30

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 $\mathbb{R}$.

• Let me add that the most reasonable assumption about $a$ is that it should be an accumulation point of the domain of the function. If a whole (punctured) interval about $a$ is contained in the domain, this is clearly true. – Siminore Oct 8 '15 at 11:59
• @Siminore I don't want to go into a long discussion, but can you just specify which scholastic tradition you are from (US, French, ..), whether you believe that a function defined only on the rationals (say $f(x)=x$) has limits and if it does whether it then is continuous on all the rationals? Possibly actually an answer would be more constructive then a comment since if you only assume acummulation point the answer to the OP's question is opposite of mine. – DRF Oct 8 '15 at 12:04
• If you say that $f: \mathbb{R} \to \mathbb{R}$ many will understand this as saying that the domain of $f$ is the set of all real numbers. – Thomas Oct 8 '15 at 12:24
• @Thomas Yes I suppose I should point out it's from the domain. – DRF Oct 8 '15 at 13:00
• Even in the US, several books does NOT require $(a-\gamma,a+\gamma)\setminus \{a\}\subset Dom(f)$. For instance see Analysis I, Serge Lang. – Ramiro Oct 8 '15 at 13:09