$f$ is continuous at $c$ $\implies$ $f$ has a limit at $c$. True?

Question

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.

Answer 1

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.