${\lim_{x\to \infty} (\frac{3x-1}{3x+1})^{4x}}$ =? ${\lim_{x\to \infty} (\frac{3x-1}{3x+1})^{4x}}$ = ?
P.S. - I tried reducing it to some form like $\lim_{n\to \infty} (1 - \frac1n)^n$, the value of which is $e^{-1}$. Was I correct?

Your idea is good and just you need some algebra:
$$\large\left(\frac{3x-1}{3x+1}\right)^{4x}=\left(1-\frac{1}{\frac{3x+1}2}\right)^{4x}=\left(\underbrace{\left(1-\frac{1}{\frac{3x+1}2}\right)^{\frac{3x+1}2}}_{\xrightarrow{x\to\infty} e^{-1}}\right)^{\frac{4x}{\frac{3x+1}2}}\xrightarrow{x\to\infty}e^{-\frac83}$$
