Does this right-handed limit exist or not? $$\lim_{x\to 0^+} \sqrt{x}$$
Schaum's Easy Outline of Calculus (Second Edition) says it does. And doesn't.
An example in the book states:
The function $f(x)=\sqrt{x}$; then $f$ is defined only to the right of zero.
Okay. So the right-handed limit does exist.
Hence, $\lim_{x\to 0} \sqrt{x}=\lim_{x\to 0^+} \sqrt{x}=0$.
Okay. I'm still with you. The limit is $0$.
Of course, $\lim_{x\to 0^+} \sqrt{x}$ does not exist,...
Qué, Mr. Fawlty?
... since $\sqrt{x}$ is not defined when $x<0$.
Okay, so are they messing with me? Is my coffee too weak? Too strong? Is there some subtle truth about limits that escapes me?
Or is that a typo? Did they mean in that last line to omit the '$+$' by the '$0$' and write "Of course, $\lim_{x\to 0} \sqrt{x}$ does not exist,..."?
EDIT:
I think a minus sign is intended instead of a plus sign in that last limit.