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To evaluate the limit

\[
\lim_{t \to 0} \frac{e^{6t} - 1}{t}
\]

we'll apply **l'Hôpital's Rule** since it results in the indeterminate form \(\frac{0}{0}\) when \(t = 0\).

### Step-by-Step Solution

1. **Differentiate the numerator and denominator**:
   - The derivative of the numerator, \(e^{6t} - 1\), is \(6e^{6t}\).
   - The derivative of the denominator, \(t\), is \(1\).

2. **Apply l'Hôpital's Rule**:
   \[
   \lim_{t \to 0} \frac{e^{6t} - 1}{t} = \lim_{t \to 0} \frac{6e^{6t}}{1} = 6e^{0} = 6.
   \]

Thus, the limit is:

\[
\lim_{t \to 0} \frac{e^{6t} - 1}{t} = 6.
\]

Would you like further explanation or have any questions on the steps?

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Here are some related questions to explore further:

1. What conditions are necessary for applying l'Hôpital's Rule?
2. How would the solution change if the exponent were \(7t\) instead of \(6t\)?
3. Can we use Taylor series to evaluate the same limit? What would the result be?
4. How does this limit relate to the derivative of \(e^{6t}\) at \(t = 0\)?
5. What are some common indeterminate forms where l'Hôpital's Rule is useful?

**Tip:** When encountering indeterminate forms like \(0/0\) or \(\infty/\infty\), l'Hôpital's Rule can simplify the problem by converting it into the limit of the derivatives.

Find the limit. Use l'Hospital's Rule where appropriate.
\[\lim_{t\to0} \frac{e^{6t} - 1}{t}\]

Find the limit of (e^(6t) - 1) / t as t approaches 0 using l'Hôpital's Rule. This problem introduces concepts of limits and exponential functions, demonstrating step-by-step application of l'Hôpital's Rule for indeterminate forms.

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