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?

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.