## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{t\to 2}(\frac{t^2-2}{t^3-3t+5})^2=\frac{4}{49}$
$\lim\limits_{t\to 2}(\frac{t^2-2}{t^3-3t+5})^2$ $=(\lim\limits_{t\to 2}\frac{t^2-2}{t^3-3t+5})^2$ (power law) $=[\frac{\lim\limits_{t\to 2}(t^2-2)}{\lim\limits_{t\to 2}(t^3-3t+5)}]^2$ (quotient law) $=[\frac{\lim\limits_{t\to 2}(t^2-2)}{A}]^2$ (1) Now we need to see if $A=0$ $A=\lim\limits_{t\to 2}(t^3-3t+5)$ $A=\lim\limits_{t\to 2}t^3-\lim\limits_{t\to 2}3t+\lim\limits_{t\to 2}5$ (difference and sum law) $A=\lim\limits_{t\to 2}t^3-3\lim\limits_{t\to 2}t+\lim\limits_{t\to 2}5$ (constant multiple law) $A=2^3-3\times2+5$ $A=7\ne0$ Therefore, (1) is defined, and quotient law is not violated. Continue with (1) $[\frac{\lim\limits_{t\to 2}(t^2-2)}{A}]^2$ $=[\frac{\lim\limits_{t\to 2}t^2-\lim\limits_{t\to 2}2}{7}]^2$ (difference law) $=[\frac{2^2-2}{7}]^2$ $=\frac{4}{49}$