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 $n+1$.


$$\lim _{ n\to \infty  } \left( \frac { n+2 }{ n-1 }  \right) ^{ 2n+3 }=\lim _{ n\to \infty  } \left( 1+\frac { 3 }{ n-1 }  \right) ^{ 2n+3 }=\\ =\lim _{ n\to \infty  }{ \left[ \left( 1+\frac { 1 }{ \frac { n-1 }{ 3 }  }  \right) ^{ \frac { n-1 }{ 3 }  } \right]  } ^{ \frac { 3 }{ n-1 } \left( 2n+3 \right)  }=\lim _{ n\rightarrow \infty  }{ { e }^{ \frac { 6n+9 }{ n-1 }  } } ={ e }^{ 6 }$$

