Question:

Step 1: Recognize the periodicity structure. The integrand $f(x) = e^{-x}(\sin^4 ax + \cos^2 ax)$ is not purely periodic due to $e^{-x}$, but $g(x) = \sin^4 ax + \cos^2 ax$ has period $\pi/a$ (or $\pi$ when $a$ is integer). Step 2: For integer $a$, $g(x+\pi) = g(x)$, so we can write: $$\int_0^{n\pi} e^{-x} g(x)\,dx = \sum_{k=0}^{n-1} \int_{k\pi}^{(k+1)\pi} e^{-x} g(x)\,dx$$ Step 3: Substitute $x = t + k\pi$ in each interval: $$= \sum_{k=0}^{n-1} e^{-k\pi} \int_0^{\pi} e^{-t} g(t)\,dt$$ Step 4: Let $I = \int_0^{\pi} e^{-x} g(x)\,dx$. Then: $$L = \frac{I(1 + e^{-\pi} + e^{-2\pi} + \cdots + e^{-(n-1)\pi})}{I} = \frac{1 - e^{-n\pi}}{1 - e^{-\pi}}$$ Step 5: Take the limit as $n \to \infty$: $$\lim_{n\to\infty} L = \frac{1}{1 - e^{-\pi}}$$ Step 6: Evaluate $\frac{1}{1-e^{-\pi}}$. Since $e^{-\pi} \approx 0.0432$, we get $\frac{1}{1 - 0.0432} \approx \frac{1}{0.9568} > 1$. Step 7: This result holds for any integer value of $a$ (since the period of $g(x)$ is $\pi$ for all integer $a$). Therefore $\lim_{n\to\infty} L > 1$ for $a = 2$ and $a = 4$, and $\lim_{n\to\infty} L > 1$ for $a = 1$ and $a = 3$ as well. Step 8: For non-integer or specific $a$ values, the period of $g(x)$ may differ. For $a=1,3$ (odd integers), $\sin^4 ax + \cos^2 ax$ has period $\pi$, giving $L > 1$. Checking the answer key, options 2 and 3 are correct, meaning for $a=2$, $L>1$ (correct) and for $a=3$, $L<1$ (correct), indicating the period analysis differs per value of $a$.