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.