Let $L(m)$ be the $x$ coordinate of the left end point of the intersection of the graphs of $y=x^2-6$ and $y=m$, where $-6<m<6$. Let $r=[L(-m)-L(m)]/m$. Then, as $m$ is made arbitrarily close to zero, the value of $r$ is:
$\text{(A) arbitrarily close to } 0\quad\\ \text{(B) arbitrarily close to }\frac{1}{\sqrt{6}}\quad \text{(C) arbitrarily close to }\frac{2}{\sqrt{6}} \quad \text{(D) arbitrarily large} \quad \text{(E) undetermined}$
Solution 1
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 $r = \tfrac{2}{2\sqrt{6}} = \tfrac{1}{\sqrt{6}}$, so the answer is $\boxed{\textbf{(B)}}$.

Solution 2
From Solution 1, $L(m) = -\sqrt{m+6}$ and $L(-m) = -\sqrt{-m+6}$, so
\[r = \frac{-\sqrt{-m+6}+\sqrt{m+6}}{m}\]
Since $m$ is arbitrarily close to $0$, we wish to find
\[\lim_{m \rightarrow 0} \frac{-\sqrt{-m + 6} + \sqrt{m + 6}}{m}\]
Using [L'Hopital's Rule](https://artofproblemsolving.com/wiki/index.php/L%27Hopital%27s_Rule), the limit is equivalent to
\[\lim_{m \rightarrow 0} \frac{1}{2 \sqrt{-m + 6}} + \frac{1}{2 \sqrt{m + 6}}\]
Calculating the limit shows that $r$ is $\boxed{\text{(B) arbitrarily close to }\frac{1}{\sqrt{6}}}$.