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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 $y = \frac{3x+1}{2}\;\;\because\;\;x \rightarrow\infty\;\therefore y \rightarrow \infty$ 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.

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