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