Answer
$0$
Work Step by Step
$\lim _{x\rightarrow -\infty }\dfrac {2x}{\left( x^{6}-1\right) ^{\frac{1}{3}}}=\dfrac {2x}{x^{2}\left( 1-\dfrac {1}{x^{6}}\right) ^{\frac {1}{3}}}=\dfrac {2}{x\left( 1-\dfrac {1}{x^{6}}\right) ^{\frac {1}{3}}}=\dfrac {2}{x\left( 1-0\right) ^{\frac {1}{3}}}=\dfrac {2}{x}=0$
