Answer
$ e^{12}$
Work Step by Step
Let $ x=\frac{n}{4}$, then we have
$$\lim _{n \rightarrow \infty}\left(1+\frac{4}{n}\right)^{3n}=\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{12x}\\
=\left(\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}\right)^12=e^{12},$$
where we used the formula $\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n}=e $.
