Answer
$e^{12}$
Work Step by Step
Given: $a_{n}=(1+\frac{3}{n})^{4n}$
We know that $\lim\limits_{n \to \infty}(1+\frac{1}{n})^{n}=e$
Assume that $\frac{n}{3}x=4n$ and solve for $x$.
we get $x=12$
and
$\frac{n}{3}\times12=4n$
Thus,
$\lim\limits_{n \to \infty}(1+\frac{3}{n})^{4n}=\lim\limits_{n \to \infty}[(1+\frac{3}{n})^{n/3}]^{12}$
Therefore, $\lim\limits_{n \to \infty}[(1+\frac{3}{n})^{n/3}]^{12}=e^{12}$
Because the limits exists , the sequence converges.
Hence, the given sequence converges to $e^{12}$.
