Does the following limit exist: $\lim _{x\to 1}\left(\sqrt{x^2-1}\right)$

Question

Should the following limit exist because if I the check the L.H.L. the function a non-real output.

$\lim _{x\to 1}\left(\sqrt{x^2-1}\right)$

Answer 1

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

Answer 2

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*}

Answer 3

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$