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The given limit is:

\[
\lim_{t \to 0} \frac{e^{4t} - 1}{\sin(t)}
\]

### Step-by-Step Solution:

1. **Direct Substitution**: 
   If we directly substitute \( t = 0 \), we get:

   \[
   \frac{e^{4(0)} - 1}{\sin(0)} = \frac{1 - 1}{0} = \frac{0}{0}
   \]

   This is an indeterminate form, so we need to use **L'Hopital's Rule** (since it is of the form \( \frac{0}{0} \)).

2. **Applying L'Hopital's Rule**:
   L'Hopital's Rule tells us to differentiate the numerator and the denominator with respect to \( t \), and then take the limit.

   - The derivative of the numerator \( e^{4t} - 1 \) with respect to \( t \) is \( 4e^{4t} \).
   - The derivative of the denominator \( \sin(t) \) with respect to \( t \) is \( \cos(t) \).

   Now, the limit becomes:

   \[
   \lim_{t \to 0} \frac{4e^{4t}}{\cos(t)}
   \]

3. **Substitute \( t = 0 \)**:
   Now, substituting \( t = 0 \):

   \[
   \frac{4e^{4(0)}}{\cos(0)} = \frac{4 \cdot 1}{1} = 4
   \]

Thus, the value of the limit is:

\[
\boxed{4}
\]

---

Do you want further details or have any questions?

Here are five related questions you might explore:
1. How do you apply L'Hopital's Rule in more complex limits?
2. What are other indeterminate forms besides \( \frac{0}{0} \)?
3. How can we compute limits without L'Hopital's Rule using series expansions?
4. What is the significance of Taylor series in finding limits involving \( e^x \) or trigonometric functions?
5. How does the behavior of exponential functions compare to trigonometric functions at limits?

**Tip**: Always check the form of the limit before applying L'Hopital's Rule—especially if it's indeterminate.

Find the limit: lim_{t → 0} (e^{4t} - 1) / sin(t)

Solve the limit lim_{t → 0} (e^{4t} - 1) / sin(t) using L'Hopital's Rule. Explore step-by-step how to handle the indeterminate form 0/0 by differentiating both the numerator and denominator. Suitable for high school calculus and college-level students.

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