1. **State the problem:** Find the limit $$\lim_{x \to \frac{\pi}{2}} \frac{3 \cos x + \cos 3x}{\left(\frac{\pi}{2} - x\right)^3}$$ 2. **Recall the formula and rules:** When evaluating limits that result in an indeterminate form like $\frac{0}{0}$, we can use series expansions or L'Hôpital's rule. Here, the denominator approaches 0 as $x \to \frac{\pi}{2}$, so we check the numerator's behavior near $x = \frac{\pi}{2}$. 3. **Rewrite the numerator using trigonometric identities:** Note that $\cos 3x = 4 \cos^3 x - 3 \cos x$. So, $$3 \cos x + \cos 3x = 3 \cos x + (4 \cos^3 x - 3 \cos x) = 4 \cos^3 x$$ 4. **Set $t = \frac{\pi}{2} - x$, so as $x \to \frac{\pi}{2}$, $t \to 0$.** Since $x = \frac{\pi}{2} - t$, then $$\cos x = \cos\left(\frac{\pi}{2} - t\right) = \sin t$$ 5. **Substitute into numerator:** $$4 \cos^3 x = 4 (\sin t)^3 = 4 \sin^3 t$$ 6. **Rewrite the limit in terms of $t$:** $$\lim_{t \to 0} \frac{4 \sin^3 t}{t^3}$$ 7. **Use the standard limit $\lim_{t \to 0} \frac{\sin t}{t} = 1$:** Therefore, $$\lim_{t \to 0} \frac{4 \sin^3 t}{t^3} = 4 \lim_{t \to 0} \left(\frac{\sin t}{t}\right)^3 = 4 \times 1^3 = 4$$ **Final answer:** $$\boxed{4}$$