limx→∞0 2x+8 2x + 4 3x-1​

$\lim_{x\to{\infty}}{\left(\dfrac{2\,x+8}{2\,x+4}\right)^{3\,x-1}}=\lim_{x\to{\infty}}{\left(1+\dfrac{4}{2\,x+4}\right)^{3\,x-1}}=\lim_{x\to{\infty}}{\left(1+\dfrac{4}{2\,x+4}\right)^{\frac{2\,x+4}{4}\,6-7}}$$\lim_{x\to{\infty}}{\left(1+\dfrac{4}{2\,x+4}\right)^{\frac{2\,x+4}{4}\,6}}\underbrace{\lim_{x\to{\infty}}{\left(1+\dfrac{4}{2\,x+4}\right)^{-7}}}_1=e^6$$