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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.

Evaluate the limit: \( \lim_{t \to 0} \frac{3^t - 2^t}{t} \)

Learn how to solve the limit \( \lim_{t \to 0} \frac{3^t - 2^t}{t} \) by applying logarithmic properties and the definition of the natural logarithm for exponential functions. This solution is ideal for students studying limits in calculus, suitable for Grades 11-12.

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