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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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    $\sqrt{x} \notin \mathbb{R}$ if x<0.2012-07-03

1 Answers 1

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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.2012-07-03