The value of $\underset{x→1}{\lim}\left[\left(\frac{4}{x^2-x^{-1}}-\frac{1-3x+x^2}{1-x^3}\right)^{-1}+3\frac{x^4-1}{x^3-x^{-1}}\right]$ is

3

$\underset{x→1}{\lim}\left[\left(\frac{4}{x^2-x^{-1}}-\frac{1-3x+x^2}{x^3-x^{-1}}\right)^{-1}+3\frac{x^4-1}{x^3-x^{-1}}\right]=\underset{x→1}{\lim}\left[\left(\frac{4x}{x^3-1}-\frac{1-3x+x^2}{1-x^3}\right)^{-1}+\frac{3x(x^4-1)}{x^4-1}\right]$

$=\underset{x→1}{\lim}\left[\left(\frac{4x+1-3x+x^2}{x^3-1}\right)^{-1}+3x\right]=\underset{x→1}{\lim}[x-1+3x]=3$