Answer
$\ln\dfrac{8}{5}$
Work Step by Step
Given: $\lim\limits_{t\to0}\dfrac{8^t-5^t}{t}$
Here, $\lim\limits_{t\to0}(8^t-5^t)=8^0-5^0=1-1=0$ and $\lim\limits_{t\to0}(t)=0$
This shows an indeterminate form of type $\dfrac{\infty}{-\infty}$, and so we will apply L'Hospital Rule.
$\lim\limits_{t\to0}\dfrac{(8^t-5^t)'}{t'}=\lim\limits_{t\to0}\dfrac{8^t\ln8-5^t\ln5}{1}$
or, $=\lim\limits_{t\to0}(8^t\ln8-5^t\ln5)$
or, $=8^0\ln8-5^0\ln5$
or,$=(1)\ln8-(1)\ln5$
or, $=\ln8-\ln5$
or, $=\ln\dfrac{8}{5}$
