Calculus: Early Transcendentals 8th Edition

$$\lim_{t\to0}\frac{8^t-5^t}{t}=\ln\frac{8}{5}$$
$$A=\lim_{t\to0}\frac{8^t-5^t}{t}$$ Since $\lim_{t\to0}(8^t-5^t)=8^0-5^0=1-1=0$ and $\lim_{t\to0}(t)=0$, so we have an indeterminate form of $\frac{0}{0}$. Now we can use L'Hospital's Rule: $$A=\lim_{t\to0}\frac{(8^t-5^t)'}{t'}$$ $$A=\lim_{t\to0}\frac{8^t\ln8-5^t\ln5}{1}$$ $$A=\lim_{t\to0}(8^t\ln8-5^t\ln5)$$ $$A=8^0\ln8-5^0\ln5$$ $$A=1\times\ln8-1\times\ln5$$ $$A=\ln8-\ln5$$ $$A=\ln\frac{8}{5}$$