#### Answer

$$\frac{1}{{{e^8}}}$$

#### Work Step by Step

$$\eqalign{
& \mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {{e^8} + h} \right) - 8}}{h} \cr
& {\text{rewrite}} \cr
& \mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {{e^8} + h} \right) - 8}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {{e^8} + h} \right) - \ln \left( {{e^8}} \right)}}{h} \cr
& {\text{using the definition of the derivative }}f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h} \cr
& {\text{comparing }}\mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {{e^8} + h} \right) - \ln \left( {{e^8}} \right)}}{h}{\text{ with the definition of the derivative}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {{e^8} + h} \right) - \ln \left( {{e^8}} \right)}}{h}:f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h} \cr
& {\text{we can note that }}f\left( x \right) = \ln x{\text{ and }}a = {e^8} \cr
& {\text{then}} \cr
& \,\,\,\,\,\,\,\,f'\left( x \right) = \frac{1}{x} \cr
& \,\,\,\,\,\,\,\,f'\left( {{e^8}} \right) = \frac{1}{{{e^9}}} \cr
& {\text{and}} \cr
& \,\,\,\,\,\,\,\mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {{e^8} + h} \right) - 8}}{h} = f'\left( {{e^8}} \right) = \frac{1}{{{e^8}}} \cr} $$