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 $n+1$.
 A: $$\begin{align}\left(\frac{n+2}{n-1}\right)^{2n+3}&= \left(1+\frac{3}{n-1}\right)^{2n+3}\\ &= \left(1+\frac{3}{n-1}\right)^{2(n-1)+5}\\ &= \left(\left(1+\frac{3}{n-1}\right)^{n-1}\right)^2\left(1+\frac{3}{n-1}\right)^5\end{align}$$
If you like, you can replace $n-1$ with $n$ as $n\rightarrow \infty$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\color{#f00}{\lim_{n \to \infty}\pars{n + 2 \over n - 1}^{2n + 3}} & =
\lim_{n \to \infty}\,\bracks{%
\pars{1 + 2/n \over 1 - 1/n}^{2n}\pars{1 + 2/n \over 1 - 1/n}^{3}}
\\[5mm] & =
\lim_{n \to \infty}\,\braces{\bracks{\pars{1 + {2 \over n}}^{n}}^{2}
\bracks{\pars{1 - {1 \over n}}^{n}}^{-2}}
\\[5mm] & =
\bracks{\lim_{n \to \infty}\pars{1 + {2 \over n}}^{n}}^{2}
\bracks{\lim_{n \to \infty}\pars{1 - {1 \over n}}^{n}}^{-2} =
\pars{\expo{2}}^{2}\pars{\expo{-1}}^{-2}
\\[5mm] & =
\color{#f00}{\expo{6}} \approx 403.4288
\end{align}
A: 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 $\log_{b}a^x = x\log_{b}a$:
$$
\log \left ( L \right ) = \lim_{x \ \rightarrow \ c} \left [ g(x) \log \left ( f(x) \right ) \right ]
$$
A: 
$$\lim _{ n\to \infty  } \left( \frac { n+2 }{ n-1 }  \right) ^{ 2n+3 }=\lim _{ n\to \infty  } \left( 1+\frac { 3 }{ n-1 }  \right) ^{ 2n+3 }=\\ =\lim _{ n\to \infty  }{ \left[ \left( 1+\frac { 1 }{ \frac { n-1 }{ 3 }  }  \right) ^{ \frac { n-1 }{ 3 }  } \right]  } ^{ \frac { 3 }{ n-1 } \left( 2n+3 \right)  }=\lim _{ n\rightarrow \infty  }{ { e }^{ \frac { 6n+9 }{ n-1 }  } } ={ e }^{ 6 }$$

A: Just another possible way to do it.
Considering $$a_n=\left(\frac{n+2}{n-1}\right)^{2n+3}$$ take logarithms $$\log(a_n)=(2n+3)\log\left(\frac{n+2}{n-1}\right)=(2n+3)\log\left(1+\frac 3{n-1}\right)$$ Now, using Taylor series for small $x$ $$\log(1+x)=x-\frac{x^2}{2}+O\left(x^3\right)$$ replace $x$ by $\frac 3{n-1}$ followed by long division to get $$\log\left(1+\frac 3{n-1}\right)=\frac{3}{n}-\frac{3}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ which makes $$\log(a_n)=6+\frac{6}{n}+O\left(\frac{1}{n^2}\right)$$ and using $a_n=e^{\log(a_n)}$ and Taylor again$$a_n=e^6+\frac{6 e^6}{n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and also how it is approached.
