How does one get from $\lim_{n \to \infty} \bigl( 1+ \frac{1}{3n} \bigr)^{2n}$ to $e^{2/3}$? Let $n \in \mathbb{N}$.
I want to know how one can find out that
$$\lim_{n \to \infty} \bigl( 1+ \frac{1}{3n} \bigr)^{2n} = e^{2/3}$$
I know that
$$\lim_{n \to \infty} \bigl(1 + \frac{x}{n} \bigr)^n = e^x$$
But which "rule" do we apply on the sequence above?
And why does this not work? If we say $\frac{1}{3n} \to 0$ with $n \to \infty$ because it's a null sequence.
Then  $\lim_{n \to \infty} (1 + 0)^{2n} = 1$ with $n \to \infty$
 A: More generally,
$\begin{array}\\
\left(1+\dfrac1{an}\right)^{bn}
&=\left(1+\dfrac1{an}\right)^{(b/a)an}\\
&=\left(\left(1+\dfrac1{an}\right)^{an}\right)^{b/a}\\
&\to e^{b/a}\\
\end{array}
$
Your case is
$a=3, b=2
$
so
$\frac{b}{a}=\frac23$
so the answer is
$e^{\frac23}
$.
A: Let $u = 3n$.  Then
$$\lim_{n \rightarrow \infty} \left( 1 + \frac{1}{3n} \right)^{2n}$$
$$= \lim_{u \rightarrow \infty} \left( 1 + \frac{1}{u} \right)^{(2/3)u}$$
$$= \left( \lim_{u \rightarrow \infty} \left( 1 + \frac{1}{u} \right)^{u} \right)^{2/3} $$
$$= e^{2/3}$$
A: If you know that: $$\lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n = e^x$$
then compare that to the limit you have. First, the exponent is $2n$, so maybe we should try a substitution of $m=2n$. As $n\to\infty$, $m\to\infty$, and vice versa, so we get:
$$\lim_{m \to \infty} \left(1 + \frac{x}{2m} \right)^{2m} = e^x$$
(And we can rename $m$ back to $n$ if we want.)
We're looking for a limit of $e^{2/3}$, so it should not come as a shock that $x=2/3$ will give us what we want:
$$e^{2/3}=\lim_{n \to \infty} \left(1 + \frac{2/3}{2n} \right)^{2n}=\lim_{n \to \infty} \left(1 + \frac{1}{3n} \right)^{2n}$$

To see why your argument that the limit is $1$ is faulty, consider this example:
(1) The limit of $n^n$ as $n\to 0$ is $0$, as we apply $n\to 0$ to the base to get $0^n$, which is $0$ (as $n\to 0$).
(2) The limit of $n^n$ as $n\to 0$ is $1$, as we apply $n\to 0$ to the exponent to get $n^0$, which is $1$ (as $n\to 0$).
But then $0=1$, which is clearly untrue! A limit cannot be equal to two different things.
The contradiction arises from not taking $n\to 0$ simultaneously for each instance of $n$ in the expression. Here, with $\lim_{n\to 0}n^n$, there is no defined limit, but in the case you give there is a limit. But the principle is the same: the variable $n$ is the same in both parts of the expression. Otherwise you are calculating:
$$ \lim_{m\to\infty}\lim_{n \to \infty} \left(1 + \frac{1}{3n} \right)^{2m} $$
or:
$$ \lim_{n\to\infty}\lim_{m \to \infty} \left(1 + \frac{1}{3n} \right)^{2m} $$
