Answer
$$-\sqrt{3}$$
Work Step by Step
We wish to find the limit
$$\lim _{t \rightarrow \infty} \tan \left(\frac{\pi 3^{t}+1}{4-3^{t+1}}\right) $$
Thus, we have:
\begin{align*}
\lim _{t \rightarrow \infty} \tan \left(\frac{\pi 3^{t}+1}{4-3^{t+1}}\right)&=\lim _{t \rightarrow \infty} \tan \left(\frac{\pi 3^{t}/3^{t}+1/3^{t}}{4/3^{t}-3^{t+1}/3^{t}}\right)\\
&=\lim _{t \rightarrow \infty} \tan \left(\frac{\pi+3^{-t}}{4 \cdot 3^{-t}-3}\right)\\
&=\tan \left(\frac{\pi}{-3}\right) \\
&=-\sqrt{3}
\end{align*}
