1. Simplifying the integral:
Using the substitution \( t = x / \epsilon \), we have \( x = \epsilon t \), so \( dx = \epsilon \, dt \).
Substituting into the integral, the limit becomes:
\[
\lim_{\epsilon \to 0} \frac{1}{\epsilon} \int_{0}^{\infty} e^{-x / \epsilon} \left( \cos(3x) + x^2 + \sqrt{x + 4} \right) dx
= \lim_{\epsilon \to 0} \int_{0}^{\infty} e^{-t} \left( \cos(3\epsilon t) + \epsilon^2 t^2 + \sqrt{\epsilon t + 4} \right) dt.
\]
2. Breaking into terms:
- For the \( \cos(3\epsilon t) \) term: As \( \epsilon \to 0 \), \( \cos(3\epsilon t) \to 1 \).
- For the \( \epsilon^2 t^2 \) term: Since \( \epsilon^2 \to 0 \), this term vanishes.
- For the \( \sqrt{\epsilon t + 4} \) term: As \( \epsilon \to 0 \), \( \sqrt{\epsilon t + 4} \to 2 \).
Substituting these limits into the integral:
\[
\int_{0}^{\infty} e^{-t} \left( 1 + 0 + 2 \right) dt = \int_{0}^{\infty} e^{-t} (3) dt.
\]
3. Evaluating the integral:
\[
\int_{0}^{\infty} 3 e^{-t} dt = 3 \int_{0}^{\infty} e^{-t} dt = 3 \left[ -e^{-t} \right]_{0}^{\infty} = 3 \left( 0 - (-1) \right) = 3.
\]
Final Answer: \( 3 \).