To solve the limit:

$\lim_{t \to 0} \frac{3^t - 2^t}{t}$

we can use the fact that for small values of $t$ approaching zero, the expressions $a^t$ (where $a$ is a constant) can be approximated using the natural logarithm, particularly using the property that

$\lim_{t \to 0} \frac{a^t - 1}{t} = \ln(a)$

Thus, we can break down the expression:

$\lim_{t \to 0} \frac{3^t - 2^t}{t} = \lim_{t \to 0} \frac{3^t - 1 + 1 - 2^t}{t}$

Separating the terms:

$= \lim_{t \to 0} \left( \frac{3^t - 1}{t} - \frac{2^t - 1}{t} \right)$

Now applying the limit property we noted above:

$= \ln(3) - \ln(2)$

Therefore, the solution is:

$\ln(3) - \ln(2)$

This can also be simplified to:

$\ln\left(\frac{3}{2}\right)$

Would you like further details on this solution or have any questions?

Here are some related questions for further exploration:

1. How does the property $\lim_{t \to 0} \frac{a^t - 1}{t} = \ln(a)$ derive from the definition of the derivative?
2. What are some applications of limits in exponential functions in real-world contexts?
3. How would the solution change if the base numbers were different, such as $4$ and $5$?
4. Can this limit be solved using L'Hôpital's Rule? Why or why not?
5. What is the intuition behind using logarithmic properties to solve exponential limits?

Tip: When handling exponential limits approaching zero, logarithmic approximations are often useful to simplify expressions.