jmc

In order for $\frac{x-2}{x^2-5}$ to be defined, we must have $x^2-5\ne 0$. So, $x\ne \pm \sqrt{5}$.

In order for $\log \frac{x-2}{x^2-5}$ to be defined, we must have $\frac{x-2}{x^2-5}>0$. There are two cases to consider: when $x^2-5>0$ and when $x^2-5<0$.

Case 1: $x^2-5>0$. Since $x^2-5>0$, we have that $x<-\sqrt{5}$ or $x>\sqrt{5}$. From $\frac{x-2}{x^2-5}>0$, we have $x-2>0$, or $x>2$. Combining all of these facts, we must have $x>\sqrt{5}$.

Case 2: $x^2-5<0$. Since $x^2-5<0$, we have that $-\sqrt{5}<x<\sqrt{5}$. From $\frac{x-2}{x^2-5}>0$, we have that $x-2<0$, or $x<2$. Combining all of these facts, we must have $-\sqrt{5}<x<2$.

So, we must have $-\sqrt{5}<x<2$ or $x>\sqrt{5}$. (We can also write this as $x=(-\sqrt{5},2)\cup (\sqrt{5},\infty )$.) The largest value not in the domain is therefore $\boxed{\sqrt{5}}$.