Let L(m) be the x-coordinate of the left endp | vedclass

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(B) The intersection points of $y = x^2 - 6$ and $y = m$ are found by setting $x^2 - 6 = m$,which gives $x^2 = m + 6$,so $x = \pm \sqrt{m + 6}$.Since

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$\frac{1}{\sqrt{6}}$

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(B) The intersection points of $y = x^2 - 6$ and $y = m$ are found by setting $x^2 - 6 = m$,which gives $x^2 = m + 6$,so $x = \pm \sqrt{m + 6}$.Since $L(m)$ is the $x$-coordinate of the left endpoint,$L(m) = -\sqrt{m + 6}$.We need to evaluate $\mathop {\lim }\limits_{m 	o 0} \left( {\frac{{L\left( { - m} ight) - L\left( m ight)}}{m}} ight)$.Substituting $L(m)$,the expression becomes $\mathop {\lim }\limits_{m 	o 0} \left( {\frac{{-\sqrt{-m + 6} - (-\sqrt{m + 6})}}{m}} ight) = \mathop {\lim }\limits_{m 	o 0} \left( {\frac{{\sqrt{m + 6} - \sqrt{6 - m}}}{m}} ight)$.Multiplying by the conjugate $\frac{\sqrt{m + 6} + \sqrt{6 - m}}{\sqrt{m + 6} + \sqrt{6 - m}}$,we get $\mathop {\lim }\limits_{m 	o 0} \left( {\frac{(m + 6) - (6 - m)}{m(\sqrt{m + 6} + \sqrt{6 - m})}} ight) = \mathop {\lim }\limits_{m 	o 0} \left( {\frac{2m}{m(\sqrt{m + 6} + \sqrt{6 - m})}} ight)$.This simplifies to $\frac{2}{\sqrt{6} + \sqrt{6}} = \frac{2}{2\sqrt{6}} = \frac{1}{\sqrt{6}}$.

Let $L(m)$ be the $x$-coordinate of the left endpoint of the intersection of the graphs of $y = x^2 - 6$ and $y = m$,where $-6 < m < 6$. The value of $\mathop {\lim }\limits_{m \to 0} \left( {\frac{{L\left( { - m} \right) - L\left( m \right)}}{m}} \right)$ equals

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