Answer
$\lim\limits_{t\to -2}\frac{t^4-2}{2t^2-3t+2}=\frac{7}{8}$
Work Step by Step
$\lim\limits_{t\to -2}\frac{t^4-2}{2t^2-3t+2}$
$=\frac{\lim\limits_{t\to -2}(t^4-2)}{\lim\limits_{t\to -2}(2t^2-3t+2)}$ (quotient law)
$=\frac{\lim\limits_{t\to -2}t^4-\lim\limits_{t\to -2}2}{\lim\limits_{t\to -2}2t^2-\lim\limits_{t\to -2}3t+\lim\limits_{t\to -2}2}$ (difference and addition law)
$=\frac{\lim\limits_{t\to -2}t^4-\lim\limits_{t\to -2}2}{2\lim\limits_{t\to -2}t^2-3\lim\limits_{t\to -2}t+\lim\limits_{t\to -2}2}$ (constant multiple law)
$=\frac{(-2)^4-2}{2\times(-2)^2-3\times(-2)+2}$
$=\frac{7}{8}$
