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.