Let $$\{x\}$$ denote the fractional part of $$x$$ and $$f(x) = \frac{\cos^{-1}(1-\{x\}^2)\sin^{-1}(1-\{x\})}{\{x\} - \{x\}^3}$$, $$x \neq 0$$. If $$L$$ and $$R$$ respectively denotes the left hand limit and the right hand limit of $$f(x)$$ at $$x = 0$$, then $$\frac{32}{\pi^2}(L^2 + R^2)$$ is equal to:

We need to find $$\frac{32}{\pi^2}(L^2 + R^2)$$ where $$L$$ and $$R$$ are the left-hand and right-hand limits of $$f(x) = \frac{\cos^{-1}(1-\{x\}^2)\sin^{-1}(1-\{x\})}{\{x\} - \{x\}^3}$$ at $$x = 0$$.

Recall that the fractional part function is defined by $$\{x\} = x - \lfloor x \rfloor$$, so as $$x \to 0^+$$ we have $$\{x\} = x \to 0^+$$ while as $$x \to 0^-$$ we have $$\{x\} = x - (-1) = x + 1 \to 1^-$$.

To compute the right-hand limit $$R$$, set $$h = \{x\} = x \to 0^+$$ so that $$f = \frac{\cos^{-1}(1 - h^2)\sin^{-1}(1 - h)}{h - h^3} = \frac{\cos^{-1}(1 - h^2)\sin^{-1}(1 - h)}{h(1 - h^2)}$$. As $$h \to 0^+$$, we have $$\cos^{-1}(1 - h^2) \approx h\sqrt{2}$$ by using $$\cos^{-1}(1-u) \approx \sqrt{2u}$$ for small $$u = h^2$$, and $$\sin^{-1}(1 - h) \to \sin^{-1}(1) = \frac{\pi}{2}$$, while $$h(1 - h^2) \to h$$. Hence $$R = \lim_{h \to 0^+} \frac{h\sqrt{2}\cdot \frac{\pi}{2}}{h} = \frac{\pi\sqrt{2}}{2} = \frac{\pi}{\sqrt{2}}$$.

For the left-hand limit $$L$$, write $$h = \{x\} = x + 1 \to 1^-$$ in the form $$h = 1 - t$$ with $$t \to 0^+$$. Then $$1 - h^2 = t(2 - t)$$, $$1 - h = t$$, and $$h(1 - h^2) = (1-t)\,t\,(2-t)$$, so that $$f = \frac{\cos^{-1}(1 - h^2)\sin^{-1}(1 - h)}{h(1 - h^2)} = \frac{\cos^{-1}(t(2-t))\sin^{-1}(t)}{(1-t)\,t\,(2-t)}$$. As $$t \to 0^+$$, we have $$\cos^{-1}(t(2-t)) \to \cos^{-1}(0) = \frac{\pi}{2}$$, $$\sin^{-1}(t) \approx t$$, and $$(1-t)\,t\,(2-t) \to 2t$$, giving $$L = \lim_{t \to 0^+} \frac{\frac{\pi}{2}\,t}{2t} = \frac{\pi}{4}$$.

Finally, since $$L^2 + R^2 = \frac{\pi^2}{16} + \frac{\pi^2}{2} = \frac{9\pi^2}{16}$$, we obtain $$\frac{32}{\pi^2}(L^2 + R^2) = \frac{32}{\pi^2} \cdot \frac{9\pi^2}{16} = 18$$.