Calculate $\lim\limits_{x\to\infty}\left(\frac{x-2}{x+2}\right)^{3x}$

Question

Calculate $\lim\limits_{x\to\infty}\left(\dfrac{x-2}{x+2}\right)^{3x}$.

I need to solve this without using L'Hôpital's rule. I tried setting $y= x+2$ to get the limit to the form: $(1-\frac{2}{y})^{3(y-2)}$ and trying to relate it to the definition of $e$, but I'm unsure where to go from there.

Answer 1

$$\left[\lim_{x\to\infty}\left(1+\frac{-4}{x+2}\right)^{\dfrac{x+2}{-4}}\right]^{\lim_{x\to\infty}\dfrac{-4\cdot3x}{x+2}}$$

Now for the inner limit use $\lim_{m\to\infty}\left(1+\dfrac1m\right)^m=e$

For the exponent $\lim_{x\to\infty}\dfrac{-4\cdot3x}{x+2}=-12\cdot\lim_{x\to\infty}\dfrac1{1+2/x}=\cdots$

Answer 2

$$\left(\frac{x-2}{x+2}\right)^{3x}=\left[\left(1-\frac4{x+2}\right)^{x+2}\right]^3\cdot\left(1-\frac4{x+2}\right)^{-6}\xrightarrow[x\to\infty]{}\left(e^{-4}\right)^3\cdot1=e^{-12}$$