<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] 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 $e^9$.
</s>