#### Answer

$\dfrac{1}{3}$

#### Work Step by Step

$\lim\limits_{t \to 2}\dfrac{t^2-4}{t^3-8}$
Factor:
$\lim\limits_{t \to 2}\dfrac{(t-2)(t+2)}{(t-2)(t^2+2t+4)}$
Simplify:
$\lim\limits_{t \to 2}\dfrac{t+2}{t^2+2t+4}$
Plug in 2:
$\dfrac{2+2}{2^2+2(2)+4}=\dfrac{4}{12}=\dfrac{1}{3}$
Therefore,
$\lim\limits_{t \to 2}\dfrac{t^2-4}{t^3-8}=\dfrac{1}{3}$