Evaluate the limits. lim _(xarrow 6^-)(x-6)/(vert x-4vert )= lim _(xarrow 6^+)(x-6)/(vert x-4vert )= Does lim _(xarrow 6)(x-6)/(vert x-4vert ) exist?

To evaluate the given limits, we need to consider the behavior of the function as $x$ approaches $6$ from both the left and the right.<br /><br />1. $\lim _{x\rightarrow 6^{-}}\frac {x-6}{\vert x-4\vert }$<br /><br />When $x$ approaches $6$ from the left, the numerator $(x-6)$ is negative, and the denominator $\vert x-4\vert$ is positive. Therefore, the fraction is negative.<br /><br />As $x$ gets closer to $6$, the numerator approaches $0$, and the denominator approaches $2$. So, the limit is:<br /><br />$\lim _{x\rightarrow 6^{-}}\frac {x-6}{\vert x-4\vert } = -\frac{1}{2}$<br /><br />2. $\lim _{x\rightarrow 6^{+}}\frac {x-6}{\vert x-4\vert }$<br /><br />When $x$ approaches $6$ from the right, the numerator $(x-6)$ is positive, and the denominator $\vert x-4\vert$ is positive. Therefore, the fraction is positive.<br /><br />As $x$ gets closer to $6$, the numerator approaches $0$, and the denominator approaches $2$. So, the limit is:<br /><br />$\lim _{x\rightarrow 6^{+}}\frac {x-6}{\vert x-4\vert } = \frac{1}{2}$<br /><br />3. Does $\lim _{x\rightarrow 6}\frac {x-6}{\vert x-4\vert }$ exist?<br /><br />For the limit to exist, the left-hand limit and the right-hand limit must be equal. However, we have:<br /><br />$\lim _{x\rightarrow 6^{-}}\frac {x-6}{\vert x-4\vert } = -\frac{1}{2}$<br /><br />$\lim _{x\rightarrow 6^{+}}\frac {x-6}{\vert x-4\vert } = \frac{1}{2}$<br /><br />Since the left-hand limit and the right-hand limit are not equal, the limit $\lim _{x\rightarrow 6}\frac {x-6}{\vert x-4\vert }$ does not exist.