Limit of $(1+3/n)^{4n}$ as $n$ goes to infinity This afternoon I was trying to evaluate $$\lim\limits_{n \to \infty} \biggl( 1 + \frac{3}{n}\biggr)^{4n}$$ 
but was having some difficulty in doing so. I know the answer to be $e^{12}$, and can easily work backwards, but I'm looking for a more insightful approach.
 A: We have that
$$
\biggl(1+\frac3n\biggr)^{4n}=\biggl(\biggl(1+\frac1{n/3}\biggr)^{n/3}\biggr)^{12}\to e^{12}
$$
as $n\to\infty$ using the continuity of the function $x\mapsto x^{12}$ for each $x\in\mathbb R$ and the fact that
$$
\biggl(1+\frac1n\biggr)^n\to e
$$
as $n\to\infty$.
A: Write as: 
$$\lim_{n\to \infty} \exp\left( 4 n\ln \left(1+\frac 3 n\right)\right)= \exp\left( 4 \lim_{n\to \infty}n\ln \left(1+\frac 3 n\right)\right)$$
$$=\exp\left(4\lim_{n\to \infty }\frac{\ln(1+3/n)}{1/n}\right)\overset{\text{LH'opital}}{=}\exp\left(4\lim_{n\to \infty} \frac{3n}{n+3}\right)=e^{4\cdot 3}=\color{red}{e^{12}}$$
A: Just another way considering $$A_n= \biggl( 1 + \frac{3}{n}\biggr)^{4n}$$ Take logarithms of both sides $$\log(A_n)=4n\log\bigl( 1 + \frac{3}{n}\bigr)$$ Now, using Taylor  series for $\log(1+x)=x-\frac{x^2}{2}+O\left(x^3\right)$, replace $x$ by $\frac 3n$ to get $$\log(A_n)=4n\biggl(\frac{3}{n}-\frac{9}{2 n^2}+O\left(\frac{1}{n^3}\right) \biggr)=12-\frac{18}{n}+O\left(\frac{1}{n^2}\right)$$ Now, $$A_n=e^{\log(A_n)}=e^{12}\biggl(1-\frac{18 }{n}+O\left(\frac{1}{n^2}\right)\biggr)$$ which shows the limit and how its approached.
