${\lim_{x\to \infty} (\frac{3x-1}{3x+1})^{4x}}$ =? ${\lim_{x\to \infty} (\frac{3x-1}{3x+1})^{4x}}$ = ?
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?
 A: Your idea is good and just you need some algebra:
$$\large\left(\frac{3x-1}{3x+1}\right)^{4x}=\left(1-\frac{1}{\frac{3x+1}2}\right)^{4x}=\left(\underbrace{\left(1-\frac{1}{\frac{3x+1}2}\right)^{\frac{3x+1}2}}_{\xrightarrow{x\to\infty} e^{-1}}\right)^{\frac{4x}{\frac{3x+1}2}}\xrightarrow{x\to\infty}e^{-\frac83}$$
A: 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.
A: As the indetermination is of the form $1^{\infty}$, use the exponential fundamental limit $\displaystyle\lim_{v\rightarrow \infty}(1+v)^{\frac{1}{v}}=e$:
$$ \lim_{x\rightarrow \infty}\left(\frac{3x-1}{3x+1}\right)^{4x}=\lim_{x\rightarrow \infty}\left(1+\frac{3x-1}{3x+1}-1\right)^{4x}=\lim_{x\rightarrow \infty}\left(1+\frac{-2}{3x+1}\right)^{4x}=$$
$$=\lim_{x\rightarrow \infty}\left(1+\frac{-2}{3x+1}\right)^{\frac{3x+1}{-2}\cdot \frac{-8x}{3x+1}}= e^{\lim_{x\rightarrow \infty}\frac{-8x}{3x+1}}=e^{-\frac{8}{3}}.$$
