Chapter 2 - Limits - 2.7 Limits at Infinity - Exercises - Page 83: 27

$\frac{1}{16}.$

We have \begin{align*} \lim _{t \rightarrow \infty}\frac{t^{4/3}+t^{1/3}}{(4t^{2/3}+1)^2} &= \lim _{t \rightarrow \infty}\frac{t^{4/3}(1+t^{-1})}{t^{4/3}(4 +t^{-2/3})^2}\\ &=\frac{(1+0)}{(4 +0)^2} \\ &=\frac{1}{16}. \end{align*}