Answer
$\lim\limits_{x \to 2}\frac{t³ + 3t² - 12t + 4}{t³ - 4t}$ = $\frac{3}{2}$
Work Step by Step
$t³ + 3t² - 12t + 4 = (t - 2)(t² + 5t - 2)$
$t³ - 4t = t(t + 2)(t - 2)$, so:
$\frac{t³ + 3t² - 12t + 4}{t³ - 4t}$ = $\frac{(t - 2)(t² + 5t - 2)}{t(t + 2)(t - 2)}$ = $\frac{(t² + 5t - 2)}{t(t + 2)}$
$\lim\limits_{x \to 2}\frac{t³ + 3t² - 12t + 4}{t³ - 4t}$ = $\lim\limits_{x \to 2}\frac{(t² + 5t - 2)}{t(t + 2)}$ = $\frac{2²+5\times2-2}{2(2+2)}$ = $\frac{12}{8}$ = $\frac{3}{2}$
