Answer $C$
Let $ L =\underset{\mathrm{x\to\infty}}{\lim}\Big(\frac{3x-1}{3x+1}\Big)^{4x} = \underset{\mathrm{x\to\infty}}{\lim}\Big(1-\frac{2}{3x+1}\Big)^{4x} = \underset{\mathrm{x\to\infty}}{\lim}\Big(1-\frac{1}{\frac{3x+1}{2}}\Big)^{4x}$
Now, Let $y = \frac{3x+1}{2}$. Since, $x \rightarrow\infty\;\therefore y \rightarrow \infty$ and $x = \frac{2y-1}{3}$
So, $$ \therefore L = \underset{\mathrm{y\to\infty}}{\lim}\Big(1-\frac{1}{y}\Big)^{4\Big (\frac{2y-1}{3}\Big )}$$
$$ = \underset{\mathrm{y\to\infty}}{\lim}\Big(1-\frac{1}{y}\Big)^{\frac{8y}{3}-\frac{4}{3}}$$
$$ = \underset{\mathrm{y\to\infty}}{\lim}\Bigg (\Big(1-\frac{1}{y}\Big)^{\frac{2y}{3}}.\Big (1-\frac{1}{y}\Big )^{\frac{-4}{3}}\Bigg )$$
$$ = \underset{\mathrm{y\to\infty}}{\lim}\Bigg (\Big(1-\frac{1}{y}\Big)^{\frac{8y}{3}}.1\Bigg )$$
$$ = \underset{\mathrm{y\to\infty}}{\lim}\Bigg (\Big(1-\frac{1}{y}\Big)^{\frac{8y}{3}}.1\Bigg )$$
$$= \Big (\underset{\mathrm{y\to\infty}}{\lim}\Big({1-\frac{1}{y}\Big )^y}\Big )^{\frac{8}{3}} \qquad \to (1)$$
But, $$ \because \underset{\mathrm{y\to\infty}}{\lim}\Big({1-\frac{1}{y}\Big )^y} = \frac{1}{e} = e^{-1}$$
So, equation $(1)$ reduces to:
$$={(e^{-1})}^{\frac{8}{3}}$$
$$= e^{\frac{-8}{3}}$$
$\therefore \; C$ is the correct option.