How to prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$? 
It's known that $\lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n = e^x$.
Using the above statement, prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$.

My attempt
Obviously, we want to reach a statement such as $$\lim_{n \to \infty}  \left(1 + \frac{-2}{n}\right)^n \quad \text{ or } \quad \lim_{n \to \infty}  \left(1 + \frac{-1}{n}\right)^n \cdot \lim_{n \to \infty} \left(1 + \frac{-1}{n}\right)^n$$
in order to be able to apply the above condition. However, I was unable to achieve this. The furthest I've got was the following:
\begin{align}
\left(\frac{n-2}{3n+1}\right)^{2n} &= \left( \frac{9n^2 - 12n + 4}{9n^2 + 6n + 1} \right)^n\\
&= \left(1 + \frac{-18n+3}{9n^2+6n+1}\right)^n\\
f(n)&= \left(1 + \frac{-2 + \frac{3}{n}}{n+\frac{2}{3} + \frac{1}{9n}} \right)^n
\end{align}
It seems quite obvious that $\lim_{n \to \infty} \left(f(n)\right) = \left(1 + \frac{-2}{n + \frac{2}{3}}\right)^n$, however, this is not exactly equal to the statement given above. Are you able to ignore the constant and apply the condition regardness? If so, why? How would you go about solving this problem?
 A: Write
$$
\frac{3n-2}{3n+1} = 1-\frac{3}{3n+1}
$$
Recall that
$$
\left(1-\frac{3}{3n+1}\right)^{3n+1} \to e^{-3}
$$
Then
$$
\begin{align}
\left(1-\frac{3}{3n+1}\right)^{3n+1}
= &\left(1-\frac{3}{3n+1}\right)^{3n} \left(1-\frac{3}{3n+1}\right)
\\ \implies&
\left(1-\frac{3}{3n+1}\right)^{3n} \to e^{-3}
\\ \implies&
\left(1-\frac{3}{3n+1}\right)^{n} \to e^{-1}
\\ \implies&
\left(1-\frac{3}{3n+1}\right)^{2n} \to e^{-2}
\end{align}
$$
A: Observe that \begin{eqnarray*}
(\frac{3n-2}{3n+1})^{2n} & = & \left(\frac{3n+1-3}{3n+1}\right)^{2n}\\
 & = & \left\{ \left(1+\frac{(-3)}{3n+1}\right)^{-1}\left(1+\frac{(-3)}{3n+1}\right)^{3n+1}\right\} ^{\frac{2}{3}}.
\end{eqnarray*}
We assume the fact without proof: For any $x\in\mathbb{R}$, $\lim_{n\rightarrow\infty}(1+\frac{x}{n})^{n}=e^{x}$.
Let $y_{n}=\left(1+\frac{(-3)}{n}\right)^{n}$. Then, 
\begin{eqnarray*}
(\frac{3n-2}{3n+1})^{2n} & = & \left\{ \left(1+\frac{(-3)}{3n+1}\right)^{-1}y_{3n+1}\right\} ^{\frac{2}{3}}.
\end{eqnarray*}
Since $y_{n}\rightarrow e^{-3}$ and $(y_{3n+1})_{n}$ is a subsequence
of $(y_{n})_{n}$, we have $y_{3n+1}\rightarrow e^{-3}$ too as $n\rightarrow\infty$.
Note that $\left(1+\frac{(-3)}{3n+1}\right)^{-1}\rightarrow1$ as
$n\rightarrow\infty$. Therefore 
\begin{eqnarray*}
\lim_{n\rightarrow\infty}(\frac{3n-2}{3n+1})^{2n} & = & \left\{ \lim_{n\rightarrow\infty}\left(1+\frac{(-3)}{3n+1}\right)^{-1}y_{3n+1}\right\} ^{\frac{2}{3}}\\
 & = & \left\{ e^{-3}\right\} ^{\frac{2}{3}}\\
 & = & e^{-2}.
\end{eqnarray*}
In the above, we have used the fact that $x\mapsto x^{\frac{2}{3}}$ is
continuous.
A: Let $m:=3n+1$. We have
$$\left(\frac{3n-2}{3n+1}\right)^{2n}=\left(1-\frac3m\right)^{2(m-1)/3}=\left(\left(1-\frac3m\right)^m\right)^{2/3}\left(1-\frac3m\right)^{-2/3}.$$
Hence the limit is $e^{-3\cdot2/3}\cdot1$.
A: We can use that
$$\left(\frac{3n-2}{3n+1}\right)^{2n}=\left(1-\frac{3}{3n+1}\right)^{2n}=\left[\left(1-\frac{3}{3n+1}\right)^{-\frac{3n+1}3}\right]^{-\frac{6n}{3n+1}}\to e^{-2}$$
A: $$L=\lim_{n \rightarrow} \left(\frac{3n-2}{3n+1} \right)^{2n}= \lim_{n \rightarrow \infty} \frac{\left([1-2/(3n)]^{3n/2}\right)^{4/3}}{\left([1+1/(3n)]^{3n}\right)^{2/3}}=\frac{ e^{-4/3}}{e^{2/3}}=e^{-2}$$
