$\lim _{x\to 1}\left(\sqrt{x-1}\right)$ I want to be clear with limits of even index root function like this one: $f(x)=\sqrt{x-1}$. The domain of this function is $[1,\infty)$.
If I'm looking for one-sided limits:
$$\lim _{x\to 1^{-}}\left(\sqrt{x-1}\right)=DNE$$ doesn't exist because the function is not defined at the left of $x=1$.
$$\lim _{x\to 1^{+}}\left(\sqrt{x-1}\right)=0$$ does exist because the function is defined at the right of $x=1$
Then,
$$\lim _{x\to 1}\left(\sqrt{x-1}\right)=DNE$$ doesn't exist because the one sided limits are not equal.
Is my analysis right?
 A: Given a real function $f:D\to\mathbb{R}$ where $D\subseteq\mathbb{R}$ we say that, for a point $x_0\in D$, $\lim\limits_{x\to x_0}f(x_0)=L$ if
$$\forall\epsilon>0\ \exists\delta>0\ \forall x\in D\cap(x_0-\delta,x_0+\delta)\setminus\{x_0\}:|f(x)-L|<\epsilon$$
that is if, zooming in sufficiently on the domain, we can make the image of every value around $x_0$ get as close as desired to the real value $L$. On the other hand, the sided limits $\lim\limits_{x\to x_0^-}f(x_0)=L$ and $\lim\limits_{x\to x_0^+}f(x_0)=L$ are respectively defined as
$$\forall\epsilon>0\ \exists\delta>0\ \forall x\in D\cap(x_0-\delta,x_0):|f(x)-L|<\epsilon$$
$$\forall\epsilon>0\ \exists\delta>0\ \forall x\in D\cap(x_0,x_0+\delta):|f(x)-L|<\epsilon$$
The thing is, in your case, since $D=[1,+\infty)$ and $x_0=1$, it turns out that the limits $\lim\limits_{x\to1^+}f(x)=\lim \limits_{x\to1}f(x)$ (and, in your case, they both exist) directly due to the fact that
$$D\cap(x_0-\delta,x_0+\delta)\setminus\{x_0\}=(1,1+\delta)= D\cap(x_0,x_0+\delta) $$
while the limit $\lim\limits_{x\to1^-}f(x)$ isn’t well determined since every point on the domain is $\geq x_0$ so $D\cap(x_0-\delta,x_0)=\emptyset$.
