The possible value(s) of k for which  $\underset{x\to \infty }{\lim }\frac{2x^3-{\left({\tan }^{-1}x\right)}^3}{\frac{8}{\pi }{x}^3{\cot }^{-1}\left|kx\right|+k^2{x}^6\sin \frac{1}{x^3}-3k{x}^3}=\frac{1}{2}$

When k=0
 $\underset{x\to \infty }{Lt}\frac{2x^3-{\left({\tan }^{-1}x\right)}^3}{\frac{8}{\pi }{x}^3.\frac{\pi }{2}+0+0}=\underset{x\to \infty }{Lt}\frac{2-{\left(\frac{{\tan }^{-1}x}{x}\right)}^3}{4}=2-1$
 $=\frac{2-0}{4}=\frac{1}{2}$
So k can be 0
If  $k\ne 0$  then  $\underset{x\to \infty }{Lt}\frac{2x^3-{\left({\tan }^{-1}x\right)}^3}{x^3\left[\frac{8}{\pi }{\cot }^{-1}\left|kx\right|+k^2\frac{\sin \frac{1}{x^3}}{\frac{1}{x^3}}-3\right]}$
 
 $\underset{x\to \infty }{Lt}\frac{2-{\left(\frac{{\tan }^{-1}x}{x}\right)}^3}{\frac{8}{\pi }{\cot }^{-1}\left|kx\right|+\frac{k^2{\sin }^{-1}\frac{1}{x^3}}{\frac{1}{x^3}}-3k}=\frac{2-0}{\frac{8}{\pi }\times 0+k^2-3k}=\frac{1}{2}$
$\begin{array}{l}\Rightarrow k^2-3k=4\\ \Rightarrow k=4,-1\end{array}$