## Calculus 8th Edition

$e^{12}$
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 $e^{12}$.