The value of $\underset{x\to 2-}{\lim }\sqrt{x^2-4}=$

Observe that $\sqrt{x^2-4}$ is not defined over $(-2,2)$

But $\underset{x\to 2+}{\lim }\sqrt{x^2-4}=0$ and $\underset{x\to 2-}{\lim }\sqrt{x^2-4}=0$

Therefore $\underset{x\to 2}{\lim }\sqrt{x^2-4}=0=\underset{x\to 2-}{\lim }\sqrt{x^2-4}$