Answer
$$\lim\limits_{t\to2}\frac{t^2-4}{t^3-8}=\frac{1}{3}$$
Work Step by Step
$$A=\lim\limits_{t\to2}\frac{t^2-4}{t^3-8}$$
- Consider the numerator:
$t^2-4=(t-2)(t+2)$
- Consider the denominator:
$t^3-8=(t-2)(t^2+2t+4)$
Therefore, $$A=\lim\limits_{t\to2}\frac{(t-2)(t+2)}{(t-2)(t^2+2t+4)}$$$$A=\lim\limits_{t\to2}\frac{t+2}{t^2+2t+4}$$$$A=\frac{2+2}{2^2+2\times2+4}$$$$A=\frac{1}{3}$$
