$$\lim_{x \to \sqrt{3}^{-}} \sqrt{x^2-3}$$
What's the answer of this limit?
There are two hyppothesis:
$0$ and undefined.
Undefined Because the square root of a negative number is undefined, $0$ because if we plug $\sqrt{3}$, we obtain $0$; I am not so sure. Please help
Edit:
Why do we calculate this limit?
According to my teacher, a function limit at sqrt(3) exists if limit at sqrt(3)- and limit at sqrt(3)+ both exists and are the same. So that's why we tried to find it.
Is that right?
Best Answer
A presumption of the usual definition of limit is that the object under consideration is a function. Note that $\sqrt{x^{2}-3}$ is not even defined if $-\sqrt{3} < x < \sqrt{3}$, so it is not legitimate to study the limit $\lim_{x \to \sqrt{3}-}\sqrt{x^{2}-3}$. However, the right-hand limit here exists and $=0$. What is the domain of the function $x \mapsto \sqrt{x^{2}-3}$?