<s>[INST] 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.
 [/INST] 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
$$\ln\left(\lim_{x \to \infty}y\right)=9 \implies \lim_{x \to \infty}y = e^9\text{.}$$
</s>