${\lim_{x\to \infty} (\frac{3x-1}{3x+1})^{4x}}$ =? ${\lim_{x\to \infty} (\frac{3x-1}{3x+1})^{4x}}$ = ?
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?

Hint:  Since all values are positive for large enough $x$,
$$
\lim_{x\rightarrow\infty}\left(\frac{3x-1}{3x+1}\right)^{4x} = \lim_{x\rightarrow\infty}\exp\left(\ln\left(\left(\frac{3x-1}{3x+1}\right)^{4x}\right)\right) = \exp\left(\lim_{x\rightarrow\infty}\ln\left(\left(\frac{3x-1}{3x+1}\right)^{4x}\right)\right) = \exp\left(\lim_{x\rightarrow\infty}4x\left(\ln\left(\frac{3x-1}{3x+1}\right)\right)\right).
$$
Now you can use L'Hospital's rule on the inner limit.
