Solution
= Solution
Let
$$
x_n=12-\frac{1000n^2+4}{1.1^n}.
$$
For $n\geq3$, the binomial theorem gives
$$
1.1^n=(1+0.1)^n
\geq\binom n3(0.1)^3
=\frac{n(n-1)(n-2)}{6000}.
$$
Consequently $(1000n^2+4)/1.1^n\to0$: given $\varepsilon>0$, the displayed rational upper bound is below $\varepsilon$ for all sufficiently large $n$. Thus, directly from the definition of convergence,
$$
\boxed{x_n\longrightarrow12}.
$$