It's a form of a something similar

When the limit is $\frac{0}{0}$ , we can use L'Hopital's Rule because the numerators and denominators of f(x) in isolation are continuous at x = 3. $\displaystyle\lim_{x \to 3} \bigg[ \frac{1}{(9 - x^2)^2} \bigg( \frac{9+x^2}{3x}- 2sin\frac{3\pi}{2}sin\frac{\pi x}{2} \bigg) \bigg] = \frac{0}{0} \\ \displaystyle\lim_{x \to 3} \bigg[ \frac{\frac{-3}{x^2} + \frac{1}{3} + \pi cos\frac{\pi x}{2}}{4x^3 - 36x} \bigg] = \frac{0}{0} \\ \displaystyle\lim_{x \to 3} \bigg[ \frac{\frac{6}{x^3} - \frac{\pi^{2}}{2} sin\frac{\pi x}{2}}{12x^2 - 36} \bigg] = \frac{\frac{6}{27} + \frac{\pi^{2}}{2}}{72} = \frac{4 + 9\pi^2}{18} \cdot \frac{1}{72} = \frac{4 + 9\pi^2}{36^2} = \frac{4 + 9\pi^2}{6^4} = \frac{a + b\pi^c}{d^a} \\ (a, b, c, d) = (4, 9, 2, 6) \Rightarrow a+b+c+d=4+9+2+6=21$