Answer
$$\lim _{t \rightarrow 2} \frac{2^{2 t}+2^{t}-20}{2^{t}-4}=9$$
Work Step by Step
Given $$\lim _{t \rightarrow 2} \frac{2^{2 t}+2^{t}-20}{2^{t}-4}$$
let $$ f(t) = \frac{2^{2 t}+2^{t}-20}{2^{t}-4}$$
Since, we have
$$ f(2)= \frac{16+4-20}{4-4}$=\frac{0}{0}$$
So, transform algebraically and cancel
\begin{aligned}
L&=\lim _{t \rightarrow 2} \frac{2^{2 t}+2^{t}-20}{2^{t}-4}\\
&= \lim _{t \rightarrow 2} \frac{(2^{ t})^2+2^{t}-20}{2^{t}-4}\\
&=\lim _{t \rightarrow 2} \frac{(2^{ t}-4)(2^t+5) }{2^{t}-4}\\
&=\lim _{t \rightarrow 2} (2^t+5) \\
&= \frac{10+1}{5+5}\\
&=2^2+5\\
&=9
\end{aligned}
