Understanding of the limit of $f\left( x\right) =\sqrt {x-1}$ as $x\rightarrow 1$ What is the limit of $f\left( x\right) =\sqrt {x-1}$ as $x\rightarrow 1$
Wade's intro.to analysis book says that ''A reasonable answer is that the limit is zero. This function, however, does not satisfy Definition3.1 because it is not an OPEN interval containing $a=1$. Indeed, $f$ is only defined for $x\geq 1$.''
Definition3.1. Let $a\in\mathbb{R}$, let $I$ be an open interval which contains $a$, and let $f$ be a real function defined everywhere on $I$ except possibly at $a$. Then $f(x)$ is said to be converge to $L$, as $x$ approaches $a$, if and only if for every $\varepsilon >0$ there is a $\delta > 0$ such that
$0 < \left| x-a\right| < \delta$ implies $\left| f\left( x\right) -L\right| < \varepsilon$.
My question is: I couldn't understand the sentence ''it is not an OPEN interval containing $a=1$.''. Why ''it is not an OPEN interval containing $a=1$''? Also, he says that  $f$ is only defined for $x\geq 1$''? Why? Why not $x\leq 1$? Can you explain clearly?

In the real numbers, only the square roots of positive numbers are defined.
$\sqrt {x-1}$ is undefined when $x< 1$
We can't find two sided limit as x approaches 1.
