Question

$\underset{t\to 8}{Lt}\frac{\sqrt[3]{3t}-2}{t^2-64}$

Answer 1

$\underset{t\to 8}{lim}=\frac{\sqrt[3]{3t-2}}{t^2-64}$ (put $t=8)\approx \frac{0}{0}$

By using $L$ Hospital Rule,

$\underset{t\to 8}{lim}=\frac{\frac{1}{3}+\frac{-2}{3}}{2t}$

$=\frac{1}{192}$