Tính \(\lim\limits_{x\rightarrow\infty}\left(\frac{3x-2}{3x+1}\right)^{2x}\).
\(e\) \(\frac{1}{e}\) \(e^2\) \(\frac{1}{e^2}\) Hướng dẫn giải:\(\lim\limits_{x\rightarrow\infty}\left(\frac{3x-2}{3x+1}\right)^{2x}=\lim\limits_{x\rightarrow\infty}\left[\left(1+\frac{-3}{3x+1}\right)^{\frac{3x+1}{-3}}\right]^{\frac{-3.2x}{3x+1}}=e^{-2}=\frac{1}{e^2}\)