Find $\lim_{x\to \infty}\left(\frac{n+2}{n-1}\right)^{2n+3}$ 
Find  $$\lim_{n\to \infty}\left(\frac{n+2}{n-1}\right)^{2n+3}.$$

My attempt:
$$\lim_{n\to \infty}\left(\frac{n+2}{n-1}\right)^{2n+3}=\lim_{n\to \infty}\left(1+\frac{3}{n-1}\right)^{2n+3}=\lim_{n\to \infty}\left(1+\frac{1}{\frac{n-1}{3}}\right)^{2n+3}$$
Now we should do something to change the power to $\frac{n-1}{3}$ because:
$\lim_{x\to \infty}(1+\frac{1}{x})^x=e$
But I cannot get the answer(the answer is $e^2$). Please give small hints not full answers.
Here is a picture from my answer:

Note that in persian $2=۲$ and $3=۳$.
edit:The answer is mistaked and take $n+1$ instad of $n+1$.

$$\begin{align}\left(\frac{n+2}{n-1}\right)^{2n+3}&= \left(1+\frac{3}{n-1}\right)^{2n+3}\\ &= \left(1+\frac{3}{n-1}\right)^{2(n-1)+5}\\ &= \left(\left(1+\frac{3}{n-1}\right)^{n-1}\right)^2\left(1+\frac{3}{n-1}\right)^5\end{align}$$
If you like, you can replace $n-1$ with $n$ as $n\rightarrow \infty$
