Let $$y = y(t)$$ be a solution of the differential equation $$\frac{dy}{dt} + \alpha y = \gamma e^{-\beta t}$$ Where, $$\alpha > 0, \beta > 0$$ and $$\gamma > 0$$. Then $$\lim_{t \to \infty} y(t)$$

Given the differential equation: $$\frac{dy}{dt} + \alpha y = \gamma e^{-\beta t}$$ where $$\alpha > 0$$, $$\beta > 0$$, and $$\gamma > 0$$.

Solve the ODE using integrating factor.

The integrating factor is $$e^{\alpha t}$$. Multiplying both sides:

$$\frac{d}{dt}(ye^{\alpha t}) = \gamma e^{(\alpha - \beta)t}$$

Case 1: $$\alpha \neq \beta$$

$$ye^{\alpha t} = \frac{\gamma}{\alpha - \beta}e^{(\alpha-\beta)t} + C$$

$$y = \frac{\gamma}{\alpha - \beta}e^{-\beta t} + Ce^{-\alpha t}$$

Case 2: $$\alpha = \beta$$

$$ye^{\alpha t} = \gamma t + C$$

$$y = (\gamma t + C)e^{-\alpha t}$$

Evaluate the limit as $$t \to \infty$$.

In Case 1: Since $$\alpha > 0$$ and $$\beta > 0$$, both $$e^{-\beta t} \to 0$$ and $$e^{-\alpha t} \to 0$$ as $$t \to \infty$$.

$$\lim_{t \to \infty} y(t) = 0$$

In Case 2: By L'Hôpital's rule, $$(\gamma t + C)e^{-\alpha t} \to 0$$ as $$t \to \infty$$ since exponential decay dominates linear growth.

$$\lim_{t \to \infty} y(t) = 0$$

In both cases, the limit is 0.

Therefore, the correct answer is Option A: $$\mathbf{\text{is } 0}$$.