# How do you find the limit lim_(x->2^+)sqrt(2-x) ?

$x \setminus \to {2}^{+}$ means that $x$ is approaching 2 from the right, that is, $x > 2$. But if $x > 2$ then $2 - x < 0$ so that $2 - x$ is not in the domain of the square root function...So $\setminus {\lim}_{x \setminus \to {2}^{+}} \setminus \sqrt{2 - x}$ does not exist , while from the left:
$\setminus {\lim}_{x \setminus \to {2}^{-}} \setminus \sqrt{2 - x} = \sqrt{0} = 0$