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First of all, assuming ONLY the knowledge of sequential characterization of limit and also the epsilon-delta formulation of limit, why is the following limit undefined?

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

This question is from Schroder's "Mathematical Analysis". My argument is following which somehow shows that for some epsilon I can always find a delta which justifies the guess that the above limit is zero:

Want: $|x|<\delta$ such that $|\sqrt{x}-0|<\epsilon$.

Argument: $|\sqrt{x}|<\epsilon$

$\leftrightarrow$ $|x|<\epsilon^2$ by squaring both sides,

Take $\delta=\epsilon^2$ for any given epsilon. Therefore the limit holds.

Anyone is welcome to point out the mistake.

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That limit makes sense only when $x\rightarrow 0$ from positive real values. –  Nirakar Neo Jul 3 '12 at 3:00
5  
I think there was a discussion about this sometime recently: under some conventions, this limit does not exist because the function is not defined on a punctured neighborhood of $0$. There are other conventions where the limit can be defined provided the limit point is an accumulation point of the domain of the function. Your book is following the former convention. The condition $0\lt |x|\lt\delta$ is not enough to guarantee that $|\sqrt{x}|\lt\epsilon$, because taking $x\lt 0$ makes the premise true, but the consequent is nonsensical. –  Arturo Magidin Jul 3 '12 at 3:02
    
$\sqrt{x} \notin \mathbb{R}$ if $x<0$. –  user29999 Jul 3 '12 at 3:03

1 Answer 1

up vote 7 down vote accepted

A common convention is to require the function to be defined in a punctured neighborhood of the point in question (but see below). The limit is undefined because the function $\sqrt{x}$ is not defined in a punctured neighborhood of $0$ (it is not defined at any negative numbers); so the limit cannot exist.

Under this convention, the definition would look as follows:

Let $a$ be a point such that $f$ is defined on a punctured neighborhood of $a$. The limit of $f(x)$ as $x\to a$ is $L$ if and only if for every $\epsilon\gt 0$ there exists $\delta\gt 0$ such that $$\text{if }0\lt |x-a|\lt\delta,\text{ then }|f(x)-L|\lt\epsilon.$$

Here, taking $x$ negative, and satisfying $|x|\lt\epsilon^2$, the premise is satisfied by the consequent is not satisfied because $f(x)$ does not even make sense.

That said: there is another convention which only requires the point to be an accumulation point of the domain. In that case, the definition would look something like:

Let $a$ be an accumulation point of the domain $D$ of $f(x)$. The limit of $f(x)$ at $a$ is $L$ if and only if for every $\epsilon\gt 0$ there exists $\delta\gt 0$ such that $$\text{if }x\in D\text{ and }0\lt|x-a|\lt\delta,\text{ then }|f(x)-L|\lt\epsilon.$$

Under that convention, your argument would be generally correct, once you add the necessary caveats that you are only considering points in the domain. But I suspect this is not the convention followed by the book you are using.

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Yes I believe what you stated was correct. –  Daniel Jul 3 '12 at 3:33

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