Here, $y_0 = y'_0 = 0$ and $f(t) = n$ for $t_0 < t < t_0+\frac{1}{n}$
Taking the Laplace transform of both sides, I've found that $$L(y) = \frac{L(f)}{As^2+Bs+C}$$ and $$L(f) = \int_0^\infty f(t)e^{-st}dt = \frac{n(e^{-st}-e^{-s(1/n+t)})}{s}$$ I'm not quite sure where to go from here, however. The question asks for the case where $n \rightarrow \infty$, does that mean that $L(f)$ simply becomes $\frac{ne^{-st}}{s}$? If so, do I just find the $y(t)$ by the inverse Laplace transform?
Note that we have the limit
$$\begin{align} \lim_{n\to \infty}\frac{n(e^{-st}-e^{-s(1/n+t)})}{s}&=\frac{e^{-st}}{s}\lim_{n\to \infty}n(1-e^{-s/n})\\\\ &=\frac{e^{-st}}{s}\times s\\\\\ &=e^{-st} \end{align}$$
Can you finish now?