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We are given the function \( f(x) = -3 + e^{-0.3x} \sin(3\pi t) \) and asked to evaluate the limit:

\[
\lim_{x \to \infty} f(x)
\]

Let's break this down:

### 1. Examine the terms:

- \( -3 \) is a constant and will not change as \( x \to \infty \).
- \( e^{-0.3x} \) decays exponentially to 0 as \( x \to \infty \) because the exponent is negative.
- \( \sin(3\pi t) \) is an oscillating function (with values ranging between -1 and 1), but its influence will diminish because it is multiplied by the exponentially decaying term \( e^{-0.3x} \).

### 2. Behavior of \( f(x) \) as \( x \to \infty \):

- The term \( e^{-0.3x} \sin(3\pi t) \) approaches 0 as \( x \to \infty \) due to the exponential decay.
- Therefore, \( f(x) \) will approach \( -3 \) as \( x \to \infty \), because the remaining non-zero term is the constant \( -3 \).

Thus, the limit is:

\[
\lim_{x \to \infty} f(x) = -3
\]

The correct answer is **-3**.

Would you like more details or have any questions?

Here are some related questions to consider:
1. How does the exponential decay impact oscillating functions in limits?
2. Why does \( e^{-0.3x} \) approach 0 as \( x \to \infty \)?
3. What role does the constant \( -3 \) play in the behavior of \( f(x) \)?
4. Can we always ignore oscillating terms when they are multiplied by exponentially decaying functions?
5. What other forms of limits might produce similar results?

**Tip:** When evaluating limits at infinity, focus on how individual terms behave—especially exponential and oscillatory components.

Let f(x) = -3 + e^(-0.3x)sin(3πt). Then lim(x -> ∞) f(x) = ?

Explore the solution to the limit problem lim(x -> ∞) f(x) where f(x) = -3 + e^(-0.3x)sin(3πt). Understand how exponential decay and oscillating functions influence the behavior of the function as x approaches infinity.

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