Chapter 2 - Limits - 2.7 Limits at Infinity - Exercises - Page 83: 28

$\frac{1}{2}.$

We have \begin{align*} \lim _{t \rightarrow \infty}\frac{t^{4/3}-9t^{1/3}}{(8t^{4}+2)^{1/3}} &= \lim _{t \rightarrow \infty}\frac{t^{4/3}(1-9t^{-1})}{t^{4/3}(8+2t^{-4})^{1/3}} \\ &=\frac{(1-0)}{(8+0)^{1/3}} \\ &=\frac{1}{2}. \end{align*}