8. - / 1 points evaluate the limit, if it exists. (if an answer does not exist, enter dne.) $lim_{h \to…

Explanation:

Step1: Rewrite negative - exponents as fractions

Rewrite ((-6 + h)^{-1}=\frac{1}{-6 + h}) and (6^{-1}=\frac{1}{6}). The limit becomes (\lim_{h\rightarrow0}\frac{\frac{1}{-6 + h}+\frac{1}{6}}{h}).

Step2: Find a common denominator for the numerator

The common denominator of (\frac{1}{-6 + h}) and (\frac{1}{6}) is (6(-6 + h)). So (\frac{1}{-6 + h}+\frac{1}{6}=\frac{6+(-6 + h)}{6(-6 + h)}=\frac{6-6 + h}{6(-6 + h)}=\frac{h}{6(-6 + h)}).

Step3: Substitute the new - form of the numerator into the limit

The limit (\lim_{h\rightarrow0}\frac{\frac{1}{-6 + h}+\frac{1}{6}}{h}) becomes (\lim_{h\rightarrow0}\frac{\frac{h}{6(-6 + h)}}{h}).

Step4: Simplify the complex - fraction

(\frac{\frac{h}{6(-6 + h)}}{h}=\frac{h}{6(-6 + h)}\cdot\frac{1}{h}=\frac{1}{6(-6 + h)}) for (h\neq0).

Step5: Evaluate the limit

Now, find (\lim_{h\rightarrow0}\frac{1}{6(-6 + h)}). Substitute (h = 0) into (\frac{1}{6(-6 + h)}), we get (\frac{1}{6\times(-6)}=-\frac{1}{36}).

Answer:

(-\frac{1}{36})