Answer
$0$
Work Step by Step
$\lim _{x\rightarrow -\infty }\dfrac {\sqrt {x^{4}-1}}{x^{3}-1}=\dfrac {\dfrac {1}{x^{3}}\sqrt {\left( x^{4}-1\right) }}{1-\dfrac {1}{x^{3}}}=\dfrac {\sqrt {\dfrac {1}{x^{6}}\left( x^{4}-1\right) }}{1-\dfrac {1}{x^{3}}}=\dfrac {\sqrt {\dfrac {1}{x^{2}}-\dfrac {1}{x^{6}}}}{1-\dfrac {1}{x^{3}}}=\dfrac {0-0}{1-0}=0$
