I am trying to figure out this limit:
$$\lim_{x \to 0^{+}} \frac{\ln{x}}{x}$$
Can I use L'Hopital's rule here? This evaluates to $\frac{-\infty}{0}$. Can you use L'Hopital's rule in this case?
Also in this example, how do I simplify:
$$\lim_{t \to 0} \frac{8^t - 5^t}{t}$$
$$ = \lim_{t \to 0} \frac{8^t \cdot \ln{8} - 5^t \cdot \ln{5}}{1}$$
Where do I go from here?
Is the derivative of $b^t$ (where b is a constant) = $b^t \cdot ln{b}$?
For $$\lim_{x \to 0^{+}} \frac{\ln{x}}{x}$$ we can not use L'Hospital rule because it is not a $\frac{0}{0}$ or $\frac{\infty}{\infty}$
For $$ \lim_{t \to 0} \frac{8^t - 5^t}{t} = \lim_{t \to 0} \frac{8^t \cdot \ln{8} - 5^t \cdot \ln{5}}{1}=\ln 8-\ln 5$$ seems right.