Does the following limit exist: $\lim _{x\to 1}\left(\sqrt{x^2-1}\right)$ Should the following limit exist because if I the check the L.H.L. the function a non-real output.
$\lim _{x\to 1}\left(\sqrt{x^2-1}\right)$
 A: 
Let's recall the definition of the limit of a real-valued function $f$ at $x=a$ as it can be found e.g. in Undergraduate Analysis by S. Lang: 
  
  
*
  
*Let $S$ be a set of numbers and let $a$ be adherent to $S$. Let $f$ be a function defined on $S$. We shall say that the limit of $f(x)$ as x approaches $a$ exists, if there exists a number $L$ having the following property.  Given $\varepsilon$, there exists a number $\delta>0$ such that for all $x\in S$ satifying $$|x-a|<\delta$$
  we have
  $$|f(x)-L|<\varepsilon.$$
  If that is the case, we write
  \begin{align*}
\lim_{{x\to a}\atop{x\in S}}f(x)=L
\end{align*}
  
  
  Thereby omitting $x\in S$ in the following examples and writing $\lim_{x\to a}f(x)=L$ whenever the domain $S$ is clear from the context.

Here we consider the function $f$
\begin{align*}
&f :[1,\infty)\to\mathbb{R}\\
&x\to f(x)=\sqrt{x^2-1}
\end{align*}

Since the domain of $f$ is $[1,\infty)$ we have the situation that in accordance with the definition of the limit above, the limit and the right-hand limit coincide. We have
  \begin{align*}
\lim_{x\to 1}\sqrt{x^2-1}=\lim_{x\to 1^+}\sqrt{x^2-1}=0
\end{align*}

A: Let's say $f(x)=\sqrt{x^2-1}$.
Now we can see that the R.H.L exists, and equal to $0$.
The problem with the L.H.L is that it is imaginary so the key question is what is the codomain of $f$?
If we say that the codomain of $f$ is including the imaginary numbers, for example, the complex plane, then the L.H.L is: $\lim_{x\to0^-}f(x)\to0i=0\leftarrow \lim_{x\to0^+}f(x)$ hence I would say that the limit exists and equal $0$.
If the codomain does not include the imaginary numbers the L.H.L is not defined, hence the limit does not exist and only the R.H.L does
A: Yes, $$\lim _{x\to 1}\left(\sqrt{x^2-1}\right)=0$$
Note that if $x>1$, then we are dealing with real numbers, and if $0<x<1$ we are dealing with complex numbers but in any case $ |\sqrt{x^2-1}|$ will approach $ 0$ 
