Answer
$e^{4/3}$
Work Step by Step
Using Theorem one, we have
$$ \lim\limits_{n \to \infty}{\frac{4n}{3n+9}}=\frac{4}{3}.$$
Now, since $e^{x}$ is continuous for all $x$, then by Theorem 4, we have
$$\lim_{n\to \infty}e^{\frac{4n}{3n+9}}=e^{ \lim_{n\to \infty}\frac{4n}{3n+9}}=e^{4/3}.$$
