Answer
$$\lim_{t\to0}\frac{8^t-5^t}{t}=\ln\frac{8}{5}$$
Work Step by Step
$$A=\lim_{t\to0}\frac{8^t-5^t}{t}$$
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:
$$A=\lim_{t\to0}\frac{(8^t-5^t)'}{t'}$$
$$A=\lim_{t\to0}\frac{8^t\ln8-5^t\ln5}{1}$$
$$A=\lim_{t\to0}(8^t\ln8-5^t\ln5)$$
$$A=8^0\ln8-5^0\ln5$$
$$A=1\times\ln8-1\times\ln5$$
$$A=\ln8-\ln5$$
$$A=\ln\frac{8}{5}$$
