Finding limit using Euler number $$\lim_{x\rightarrow \infty }\left ( 1+\frac{3}{x+2} \right )^{3x-6}$$
I've tried to factor and simplfy the expression. I got:
$${\left ( 1+\frac{3}{x+2} \right )^{\frac{1}{x+2}}}^{3({x^2-4})}$$
I set $x$ to $1/t$ I get:
$${\left ( 1+\frac{3}{\frac{1}{t}+2} \right )^{\frac{1}{\frac{1}{t}+2}}}^{3 \left({\frac{1}{t}^2-4} \right)}$$
then I am left with:
$$\left ( e^{3} \right )^{3\left(\frac{1}{t^2}-4\right)}$$ which I get by using Euler number.
The answer is $e^9$, but clearly the answer I get is $(e^9)^{\text{expression}}$ which is not equal to the answer.

$\lim_\limits{x\to\infty}\left(1+\frac {3}{x+2}\right)^{3x-6}$
$y = x+2$
$\lim_\limits{y\to\infty}\left(1+\frac {3}{y}\right)^{3y-12}\\
\lim_\limits{y\to\infty}\left(1+\frac {3}{y}\right)^{3y}\lim_\limits{y\to\infty}\left(1+\frac {3}{y}\right)^{-12}$
Let't attack these separately
$\lim_\limits{y\to\infty}\left(1+\frac {3}{y}\right)^{-12} = 1$
As $y$ gets to be large $\frac {3}{y}$ becomes effectively $0,$ and the limit goes to 1.
$\lim_\limits{y\to\infty}\left(\left(1+\frac {3}{y}\right)^{y}\right)^3\\
\lim_\limits{y\to\infty}\left(1+\frac {3}{y}\right)^{y} = e^3$
$\lim_\limits{y\to\infty}\left(\left(1+\frac {3}{y}\right)^{y}\right)^3 = (e^3)^3 = e^9$
