Solution
= Solution
For fixed $x>0$, $f_n(x)\to e^{-x}\sin x$; for $x\le0$, the limit is zero. On $0<x<n$, the supplied inequality gives $(1-x/n)^n\le e^{-x}$, so
$$
|f_n(x)|\le\chi_{(0,\infty)}(x)e^{-x}.
$$
This is the given integrable dominating function. By the <dominated convergence theorem>,
$$
\lim_n\int f_n(x)\,dx=\int_0^\infty e^{-x}\sin x\,dx
=\operatorname{Im}\frac1{1-i}=\boxed{\frac12}.
$$