Answer
$\lim\limits_{t\to 2}(\frac{t^2-2}{t^3-3t+5})^2=\frac{4}{49}$
Work Step by Step
$\lim\limits_{t\to 2}(\frac{t^2-2}{t^3-3t+5})^2$
$=(\lim\limits_{t\to 2}\frac{t^2-2}{t^3-3t+5})^2$ (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)
$=[\frac{2^2-2}{7}]^2$
$=\frac{4}{49}$
