Let $\left\{x\right\}$ denote the fractional part of x and $f\left(x\right)=\frac{{\cos }^{-1}(1-{\left\{x\right\}}^2){\sin }^{-1}(1-\{x\left\}\right)}{\left\{x\right\}-{\left\{x\right\}}^3}, x\ne 0.$ If L and R respectively denotes the left hand limit and the right hand limit of f(x) at x = 0, then $\frac{16}{{\pi }^2}(L^2+R^2)$ is equal to_______.

Finding right hand limit
$\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f(0+h)=\underset{h\to 0}{\lim }f\left(h\right)$ 
$=\underset{h\to 0}{\lim }\frac{{\cos }^{-1}(1-h^2){\sin }^{-1}(1-h)}{h(1-h^2)}=\underset{h\to 0}{\lim }\frac{{\cos }^{-1}(1-h^2)}{h}\left(\frac{{\sin }^{-1}1}{1}\right)$ 
Let ${\cos }^{-1}(1-h^2)=\theta \Rightarrow \cos \theta =1-h^2$  
$=\frac{\pi }{2}\underset{\theta \to 0}{\lim }\frac{\theta }{\sqrt{1-\cos \theta }}=\frac{\pi }{2}\underset{\theta \to 0}{\lim }\frac{1}{\sqrt{\frac{1-\cos \theta }{{\theta }^2}}}=\frac{\pi }{2}\frac{1}{\sqrt{1/2}}$ 
$R=\frac{\pi }{\sqrt{2}}$ 
Now finding left hand limit
$L=\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f(-h)$ 
$=\underset{h\to 0}{\lim }\frac{{\cos }^{-1}(1-{\{-h\}}^2){\sin }^{-1}(1-\{-h\left\}\right)}{\{-h\}-{\{-h\}}^3}=\underset{h\to 0}{\lim }\frac{{\cos }^{-1}(1-{(-h+1)}^2){\sin }^{-1}(1-(-h+1\left)\right)}{(-h+1)-{(-h+1)}^3}$ 
$=\underset{h\to 0}{\lim }\frac{{\cos }^{-1}(-h^2+2h){\sin }^{-1}h}{(1-h)(1-{(1-h)}^2)}=\underset{h\to 0}{\lim }\left(\frac{\pi }{2}\right)\frac{{\sin }^{-1}h}{(1-{(1-h)}^2)}$ 
$=\frac{\pi }{2}\underset{h\to 0}{\lim }\left(\frac{{\sin }^{-1}h}{-h^2+2h}\right)=\frac{\pi }{2}\underset{h\to 0}{\lim }\left(\frac{{\sin }^{-1}h}{h}\right)\left(\frac{1}{-h+2}\right)$ 
$L=\frac{\pi }{4}$ 
$\frac{16}{{\pi }^2}(L^2+R^2)=\frac{16}{{\pi }^2}(\frac{{\pi }^2}{2}+\frac{{\pi }^2}{16})=9$