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.
 A: To get the result notice that $3x-6 = 9 \cdot \frac{x+2}{3} - 12$ then you limit is
$$
\lim_{x \to +\infty} \left[ \left( 1 + \frac{1}{\frac{x+2}{3}} \right)^{\frac{x+2}{3}} \right]^9 \cdot \left( 1 + \frac{3}{x+2} \right)^{-12}.
$$
The second term goes to $1$, while the term inside the brackets goes to $e$. Therefore, you get that the limit is equal to $e^9$.
A: $\lim_\limits{x\to\infty}\left(1+\frac {3}{x+2}\right)^{3x-6}$
$y = x+2$
$\lim_\limits{y\to\infty}\left(1+\frac {3}{y}\right)^{3y-12}\\
\lim_\limits{y\to\infty}\left(1+\frac {3}{y}\right)^{3y}\lim_\limits{y\to\infty}\left(1+\frac {3}{y}\right)^{-12}$
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.
$\lim_\limits{y\to\infty}\left(\left(1+\frac {3}{y}\right)^{y}\right)^3\\
\lim_\limits{y\to\infty}\left(1+\frac {3}{y}\right)^{y} = e^3$
$\lim_\limits{y\to\infty}\left(\left(1+\frac {3}{y}\right)^{y}\right)^3 = (e^3)^3 = e^9$
A: Let $y = \left(1+\dfrac{3}{x+2}\right)^{3x-6} $. Then 
$$\ln y=(3x-6)\ln\left(1+\dfrac{3}{x+2}\right)\text{.}$$
On the left-hand side, by continuity of $\ln$, 
$$\lim_{x \to \infty}\ln y = \ln \left(\lim_{x \to \infty}y\right)\text{.}$$
On the right-hand side,
$$\lim_{x \to \infty}(3x-6)\ln[1+3/(x+2)]=\lim_{x \to \infty}\dfrac{\ln[1+3/(x+2)]}{1/(3x-6)}\text{.}$$
Apply L-Hospital's rule. The numerator changes to
$$\dfrac{1}{1+3/(x+2)} \cdot \dfrac{-3}{x^2}$$
and the denominator changes to
$$\dfrac{-1}{3x^2}$$
and furthermore, 
we have
$$\lim_{x \to \infty}\dfrac{1/[1+3/(x+2)] \cdot -3/x^2}{-1/(3x^2)} = \lim_{x \to \infty}\dfrac{9}{1+3/(x+2)} = 9\text{.}$$
By L-Hospital,
$$\lim_{x \to \infty}\dfrac{\ln[1+3/(x+2)]}{1/(3x-6)} = 9\text{.}$$
Hence, taking into account the left-hand side, we have
$$\ln\left(\lim_{x \to \infty}y\right)=9 \implies \lim_{x \to \infty}y = e^9\text{.}$$
