10・88 Let $L(m)$ be the $x$-coordinate of the left intersection point of the graphs of $y=x^{2}-6$ and $y=m$ (where $-6<m<6$). Let $r=\frac{L(-m)-L(m)}{m}$, then, as $m$ approaches 0, the value of $r$ is
(A) very close to 0.
(B) very close to $\frac{1}{\sqrt{6}}$.
(C) very close to $\frac{2}{\sqrt{6}}$.
(D) infinitely large.
(E) cannot be determined.
(20th American High School Mathematics Examination, 1969)
[Solution] Suppose the coordinates of $L(m)$ satisfy $x^{2}-6=m$ or $x^{2}=6+m$. That is, $x= \pm \sqrt{6+m}$, which is not 0 (note that $-6<m<6$).
Therefore, its left endpoint is $L(m)=-\sqrt{6+m}$, so
$$
r=\frac{L(-m)-L(m)}{m}=\frac{-\sqrt{6-m}-(-\sqrt{6+m})}{m} \text {, }
$$

which means $\square$
$$
r=\frac{2}{\sqrt{6+m}+\sqrt{6-m}} .
$$

Then, as $m \rightarrow 0$, $r \rightarrow \frac{2}{\sqrt{6}+\sqrt{6}}=\frac{2}{2 \sqrt{6}}=\frac{1}{\sqrt{6}}$. Therefore, the answer is $(B)$.