Answer
$$ \sqrt{3}$$
Work Step by Step
Given $$\lim _{x \rightarrow - \infty}\frac{\sqrt { 3x^4+x}}{\sqrt {x^2-8} } $$
Then
\begin{align*}
\lim _{x \rightarrow - \infty}\frac{\sqrt { 3x^4+x}}{\sqrt {x^2-8} } &=\lim _{x \rightarrow - \infty}\frac{\sqrt { 3x^4/x^4+x/x^4}}{\sqrt {x^2/x^2-8/x^2} } \\ &=\lim _{x \rightarrow - \infty}\frac{\sqrt { 3 +1/x^3}}{\sqrt {1-8/x^2} } \\ &= \sqrt{3} \end{align*}