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Further to Another simple/conceptual limit question where I was questioning David Brannan's assertion in his A First Course in Mathematical Analysis that $f(x)=\sqrt x,x\geq 0$ has no limit at $0$ (Example 2c if you type in page 184 in the box on http://www.scribd.com/doc/74564079/Mathematical-Analysis), I just noticed that in an earlier section he asserted that the same function is continuous on its domain i.e. including at $0$ (Example 3 if you type in page 148 in the box).

Does this not violate the well-known theorem (Thm 2 if you type in page 185 in the box) implying that if $f$ is continuous at $c$, then $f$ has a limit at $c$ ? Is David Brannan contradicting himself (by setting up his definition of limits badly)?

EDIT: Thanks, Wisefool. It turns out that there is no contradiction in Brannan's assertions (that the square root function is continuous at 0 yet has no limit at 0) after all, but only because of his peculiar definition of "limit" and his analogously peculiar statement of the theorem relating continuity at a point to the limit at that point.

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    $\begingroup$ I fail to see the problem. The function you mention is continuous on its domain and it does have a limit at x=0. $\endgroup$
    – Ittay Weiss
    Jan 23, 2013 at 7:22
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    $\begingroup$ @IttayWeiss. What I understood the problem here is, is that the author of the book claims that there is no limit for $\sqrt{x}$ in $0$, while in other place he claims that for a continuous function the limit exists in every point of its domain. $\endgroup$
    – T. Eskin
    Jan 23, 2013 at 7:25
  • $\begingroup$ Yes, Thomas, that's exactly the issue- thanks. $\endgroup$
    – ryang
    Jan 23, 2013 at 7:32
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    $\begingroup$ IMO this is exactly the kind of trouble you get into when you try to define functions like $\sqrt{x}$ over all of $\mathbb{R}$ and start talking about "defined" and "undefined" points, instead of making full use of topology and subspaces. $\endgroup$
    – user7530
    Jan 23, 2013 at 7:50
  • $\begingroup$ @user7530 It's simply David Brannan being clumsy in dumbing down the definition of limit (not letting the limit to exist at $c$ when $c$ is an end-point of $f$'s domain) to suit whatever purpose he has. But I agree that the trouble run into is the same. $\endgroup$
    – ryang
    Jan 23, 2013 at 8:46

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It is a simple matter of terminology: in that example (or in that section/chapter/book) a limit is understood to be a two-sided limit. $f$ has limit $l$ in $x_0$ if:

  1. $f$ is defined in a (puntcured) neighborhood of $x_0$
  2. for every $\epsilon>0$ there exists $\delta>0$ such that if $|x-x_0|<\delta$ then $|f(x)-l|<\epsilon$.

In that example, it is the first condition which fails.

This isn't exactly standard and personally I would have said that the function had a limit for $x\to 0$, but I can see the book's point...

EDIT: moreover, Thm2 on page 185 speaks about a point $c$ contained in an open interval $I$ where the function is defined. There is no open interval of the real line containing $0$ where the function $\sqrt{x}$ is defined.

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  • $\begingroup$ Thanks for pointing out that Thm 2 mandates that the point in question isn't an end-point. I had missed that. $\endgroup$
    – ryang
    Jan 23, 2013 at 7:34

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