smc

Since $L(a)$ is the $x$ coordinate of the left end point of the intersection of the graphs of $y=x^2-6$ and $y=a$, we can substitute $a$ for $y$ and find the lowest solution $x$. $$x^2-6=a$$ $$x=\pm \sqrt{a+6}$$ That means $L(m)=-\sqrt{m+6}$ and $L(-m)=-\sqrt{-m+6}$. That means $$r=\frac{-\sqrt{-m+6}+\sqrt{m+6}}{m}$$ Since plugging in $0$ for $m$ results in $0/0$, there is a removable discontinuity. Multiply the fraction by $\frac{\sqrt{-m+6}+\sqrt{m+6}}{\sqrt{-m+6}+\sqrt{m+6}}$ to get $$r=\frac{m+6-(-m+6)}{m(\sqrt{-m+6}+\sqrt{m+6})}$$ $$r=\frac{2m}{m(\sqrt{-m+6}+\sqrt{m+6})}$$ $$r=\frac{2}{(\sqrt{-m+6}+\sqrt{m+6})}$$ Now there wouldn't be a problem plugging in $0$ for $m$. Doing so results in $r=\frac{2}{2\sqrt{6}}=\frac{1}{\sqrt{6}}$, so the answer is $\boxed{\text{arbitrarily close to}\frac{1}{\sqrt{6}}}$.