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 $n+1$.

Hint:
$$
L = \lim_{x \ \rightarrow \ c} f(x)^{g(x)} \implies \log \left ( L \right ) = \log \left ( \lim_{x \ \rightarrow \ c} f(x)^{g(x)} \right )
$$
Then by the continuity of $\log(x)$ and using $\log_{b}a^x = x\log_{b}a$:
$$
\log \left ( L \right ) = \lim_{x \ \rightarrow \ c} \left [ g(x) \log \left ( f(x) \right ) \right ]
$$
