## Calculus: Early Transcendentals 8th Edition

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